hello everyone in this video we're going to review the ab pre-calculus topic 1.12 1.6 all right question number one the graph of the polinomial functions f is shown above where 5 less than equal to S less than equal to 5 so which mean means this is the domain of the function right so the function stop right here the function f has local s at s² -2 and s² 2 so this is local maximum this gu is local minimum right and the graph of f has a point of inflection at s = to zero so here's the point of inflection right here all right on what intervals is f increasing all right so in order to know if the function is increasing or decreasing make sure you know the Christmas tree trick okay so I have the Christmas tree trick right here okay so you need to draw the tangent lines of the graph right start the tangent line like this this the first tangent line this is the second second tangent line and this a third tangent line now if you see the graphs going this way the tangent lines going this way the slope is positive if the slope positive the function is increasing right if the L going this way the slope is negative and if the slope is negative the function is decreasing all right so look at this one here the function is increasing from -5 to -2 right here right from here to here all right so it's going to be5 2 -2 Union this guy right here all right net 2 two five right on quad intervals is the graph of f is concave down okay so if you see the graph go like this this concave down this gu concave up right so this is down and this is concave up oh you see that if you have half of a concave T like this or like this this a con con cave down this guy like this or like this concave up right so we look at this one it's going to be from 5 to 1 right here all right so if you look at this one it's going to be from 5 to Zer right here right the function is Con down so it could be5 comma Z all right number two right here I'm going to leave it for you to practice now let's move on to question three the graph of H is shown above has relative ASA s s -3 and S 2 2 so s -3 we have the local minimum right and S = to 2 we have the local maximum right and a point of inflection s s = to ne1 so s negative one the point of inflation right here all right what intervals is hedge increasing and concave down increasing and concave down so again you're going to draw the tangent line right you're going to draw this guy this tangent line This tangent line and this tangent line right here now increasing which means the slope is positive so only this guy and this guy right here the slope is positive now if you have the tent like right here the slope is positive but the function is concave up right this guy the slope is positive which means the function is increasing but it's concave down right so it's going to be from -1 to two right here so it could be1 2 2 now question B on what intervals is H decreasing and and concave up all right decreasing and concave up so we look at this one this guy tells us the function is decreasing right and this guy right here decreasing right but concave up this guy is concave up this guy is concave down right so decreasing and concave up should be this guy so it's could be negative infinitive to3 right here so it could be negative infinitive 2 -3 now on what intervals each H decreasing at a decreasing rate all right the function is decreasing so we look at this guy right here either this guy right or this guy now add a decreasing range so you're going to draw two tension Lin and you compare right so right here I have the first tangent line right here now I have the second tangent line it's going be right here all right use the Christmas Tre trick again all right if the tension lines going this way the slope is positive the tension lines going this way the slope is negative all right if the line going this way and if the line is ther the slope is bigger right so T line is ther the slope is biger now if the line go this way and then if the tension line is steeper the slope is smaller now we look at this one we have two tensent lines right here and then the function is decreasing right and then you look at this one right here this guy right here the first one right here is steeper than the second one so it's going to be this one stipper the slope is smaller so which means this guy increasing right the r is increasing right but we need decreasing in R so this is not the case now look at this one right here we draw another tangent light right here all right and we see that this guy right here stipper than this guy right so this guy right here has a smaller slope than this guy which means the slope is decreasing right so the answer must be two to infinitive you're going to put infinitive right here positive Infinity right here and negative Infinity Play here now 3D on what intervals each Edge increasing at an increasing rate now we look at this one right here so we have two increasing interval right here from 3 to1 and from1 to two right so we're going to explore each of them right now first one I'm going to draw another tangent line right here I see that the tangent line is going this way which mean the function is increasing so I'm good with this one right now I'm going to check if the rate of chain is increasing or decreasing so I look at this one right here I see that the tension line is getting stipper which means the slope is bigger so the function has an increasing rate so the answer must be this interval so it's could be -3 21 all right and if you do this one the function is increasing at a decreasing rate from1 to two right so we only have one answer right here -3 21 me determine if each of the following statement is true or false now question four the rate of change of the function case is increasing over the interval 4 to six so we look at this one 4 to six this is four and this is six right here so this guy right here yes the rate of change of case is increasing yes you can draw two tangent line and you see that the first tangent line right here and the second tangent line right here and you see that the second tangent L is deeper than the first one which means the slope is bigger so the rate of change is increasing right so this is true now question number five the rate of change of the function K is negative over the interval -4 to0 -42 Z which means this guy right here all right so the rate of chain of the function K is negative yes you can draw a tangent line and you see that tangent line is going this way which means we have the Christmas tree trick right here so the rate of change of K is negative over this interval so this is true now question six the function K is in increasing at a decreasing rate over the interval 2 to 4 so 2 to 4 we have this guy and to this guy right here all right so we draw one tangent line and we see that the tangent line going this way which mean the function is decreasing but this guy say the function is increasing so this wrong so this is super false now question number seven the rate of change of the function K is positive and decreasing over the interval 6 to4 so6 right here4 right here we have the function is going be right here all right so we're going to check the r CH of K is positive so we draw the tangent line and we see that the tangent line going this way which means the rate of chain is positive which is good this is right and decreasing right we draw another tangent line so it could be right here and we see that the tensent line getting less step which means it has smaller slope and then the function has a decreasing grate right so this is true so question number seven is true now question eight the graph of the function K is concave down over the interval -4 to4 so -4 to 4 is going to be right here to this one right here all right so this is not the case because this guy right here from here to here the function is concave up and from here to here the function is concave up so that's wrong so number eight is false question number nine the function K is increasing over the interval 6 to4 ne6 right here to -4 right here so this -6 this4 the function is increasing yes because we can draw a tangent line and we see that tent L going this way right here so the slope is positive which means the function is increasing from negative 6 to4 now check 4 to6 4 to6 it be from here to here and we draw another tent line and then yes the function has the tensent line has the positive slope so the function is increasing so number nine is true now question 10 the function K is negative and increasing over foris negative and increasing over foris sub foris right here right uh the function K is negative so we check this one we see that the function is below the F axis so this is negative so this is true right and increasing right which draw another the tangent line we see that the tangent L going this way so which means it has the positive slope so which means the function is increasing right so question 10 this is true now question 11 the function K is decreasing at a decreasing rate over the interval -320 and 22 4 right so -320 we going to draw this one -320 it's going be right here and 224 224 is going to be right here all right now we're going to check if the function is decreasing at a decreasing rate over these intervals so we draw the tangent line and we see that tangent line going this way which mean the function is decreasing good all right we draw another tangent line right here and another tensent line right here and we see that the slope getting bigger because the tension lines going this way and then if the tension line stipper the slope is smaller right now this guy right here less steep than this guy which means the slope is bigger so the function is decreasing at an increasing rate but it say the function is decreasing and decreasing R so this is false number 12 the function K is positive and the graph of K is concave up over the interval -3 3 to Z -3 to Z so we have this guy right here all right the function is positive yes because it's above the xaxis so the function is positive and the graph of K is concave up yes this is concave up right we have like something like this or something like this this is concave up so the answer number 12 is true now number 13 the function is positive and the rate of CH of taste is positive over the interval to 4 to six so 4 to six is going to be right here all right so the function is positive that's wrong right because the function is below the xaxis so this negative so this is false all right so we know that the change formula is going to be F of B minus F of A over B minus a over the interval a b right so we look at the first one right here number 14 find the average rate of change of the function f ofs over this interval right here so we have average rate of CH is going to be F of 2 minus F of 1/ 2 - 1 so we just substitute into the function right here so we have 2 * 2 Q - 2 + 1 all right don't forget parenthesis minus another parenthesis so it could be 2 * 1 Q - 1 + 1 over one right so if you simplify this one it turns out to be 13 now question number 15 is going to be similar to the question number 14 so I'm going to leave it for you okay now question number 16 find the average gra change of function G of X over the interval 1 to 4 all right the function G of X this is the P function all right all right so now we're going to use the formula it's going to be aage rate of is f of 4 - F of -1 over 4 -1 so it turns out to be F four it's could be because four greater than three so we're going to use the second one three here so it's going to be 4 squ + 2 all together subtract 1 Which is less than three right here so we're going to use the first one right here so it could be 4 minus 2 * -1 over 4 -1 should be F right so if we do this one and we simplify this one it turns out to be 12 over 5 right now question number 17 is similar to question number 16 I leave it for you okay now question number 18 find the average R of ch now question number 17 and 18 similar to question number 16 right so I just leave it for you now question number 19 find the average rate of chance of the function K ofs Over the interval negative 2 to one right so we have the average rate of chain is K of one minus K of -2 over 1 minus -2 so it's going to be K look at the function K right here K of 1 is going to be 4 minus K of -2 is going to be 8 over 3 right so it's could be negative 4 over three right now look at number 20 find the average change of the function P of s over the interval -125 so -125 is going to be right here all right so we have the a change formula is P of 5 minus p of1 over 5 -1 so P of five we look at this one we see that it's going to be three right so this guy is going to be 3 minus P of 1 p of1 is going to be one right here over 5 - 1 is going be six so it could be 13 now number 21 I leave it for you okay number 22 to number 27 determine if the following functions could be concave up concave down or neither in order to do this one we need to calculate the average rate of change if the average a rate of change is increasing the function is concave up so average rate of change increasing the function is concave up if average rate of CH is decreasing the function is concave down and if average rate of change both increasing and decreasing the function is neither all right so we look at this one we see that now we need to find the change of this one so does it mean in order to get four from three I'm going to add one right this one I'm going to add two this one I'm going to add three and this one I'm going to add six right so in order to find the average rate of change I'm going to take one and divide by the equalent in both intervals right here this one one this one one this one one so the a CH the first one is going be 1 over one the second one 2 over one the third one 3 over one the last one sit over one and you see that average of chance is increasing right so because average of change is increasing the function is Con up right in order to get one from -3 I'm going to add four and this one I'm going to add six I'm going to add four and I'm going to add four so so the aage rate of change can be 4 over 2 right because the equal length input interval is could be two right so it could be two now you have another interval right here 1 27 right here so app change going be six over two and it's going to be three and this one right here is going to be four over two and the last one here 4/ two so you see that right right at in this table the function is is about have the aate chance increasing and constant at the same time so it should be nther seven and this one five this one two this one one right so the aate change going be 7 over three 5 over 3 2 over three and 1 over three and you see that the average rate of change is decreasing right so the function is concave there all right so I'm going to leave number 25 26 and 27 for you right now let's take a look at the 28 to 31 determine the least possible degree of the pols right so you justy the change in y over the equivalent input intervals right so we look at this one right here in order to get -3 from3 you just add zero right this guy right here you just add two four this guy right here you are seven this guy right here you add nine and this guy right here you add 10 all right so this output are not the same so you keep calculating right so in order to get four from youro I'm going to add four this guy I'm going to add three this guy I'm going to add two and this one I'm going to add one they are not the same right so next in order to get three from four I'm going to subtract one this one I'm going to subtract one this one I'm going to subtract one so after the first second and the third difference all the are put right here at the same right so the function has the degree three because the thir difference are constant now look at number 29 over the equalent input intervals right here we're going to calculate the difference of the output so in order to get eight from 12 I'm going to subtract four subtract two now look at number 30 right over the equivalent input interval we have the change in the output right so this one subtract five subtract two so + one + 5 and + 11 right next one we're going to add three this one we're going to add now we're going to add zero one two all right add one and then add one so after one 2 3 four times calculate the difference at the output we have the same constant right here so the function has decree four 1 is could be three is could be five is could to be seven and it's going to be nine right the chain in output is going to be 1 3 5 7 and 9 now we're going to calculate the average R chain so okay 1/ one is going to be one the second one 3 over three is going to be one the third one 5 over five is going to be one the next one 7 over 7 is going to be one and the last one 9 over 9 is going to be one so because the average rate of change is a constant so the function is a linear right so this is degree 1 now let's move on to question number 32 to 38 so question number 32 to 38 determine if the following functions are even not or nether so keep that in mind we have the formula if f ofx is f of x the function is R if you have f ofx is f of x the function is even right so we're going to check this one so we have y equal this one right so I just say F of so I just name this function as f of x right so I'm going to check f of x is going be 2 * right when we see s i subtitute s right so I have s q minus 3 * s + 9 right I'm going to simplify s is going be S so could be -2 s CU -3 * s be + 3 n + 9 right so because this one right here is not the same thing as original one right no it's not going to be negative f ofx so this one should be nether now question number 34 we start from f of negative s is ative s to the 4 minus 3 make sure you put the parenthesis so can be because this one has the even power so any negative number raised to the even power is going to be positive so we have S to the 4 minus 3 so let go back to original one right here right so we have this one is going to be f of x so we just prove that f of x equals f of x so the function is even right now the last one right here we have G of s is going to be -6 * X all together ra to the power 5 + 2 * s CU minus s right here right so this guy turns out to be negative s² to the fth all right so it's going to be S because this is our power so s is turn out to be S to the 5th right so s to the 5th multip by6 so it could be 6 s to the 5th this guy -2 s CU and this one plus s now I can Factor the negative side out so it could be what negative side I put this outside so this guy turn out to be 6 s to the 5th plus 2 s CU minus s right so you see the inside right here this is actually the function G of s right so I rewrite the negative side outside this guy actually actually the function G of s right here so I just put that g of s equal G of s so the function is out now look at number 36 keep in mind the function is even EXC symmetrical about the Y AIS if the function is out it's going to be symmetrical about the origin all right the point right here it's going to be four three so because we have the opposite point right here is going to be 4 and -3 right so which means the function symmetric about the origin right here so the function is R we look at number 37 right here right the function is not symmetric about the origin not the y axis so this is neither number 38 the function symmetrical about the y axis rate look if we have the P here we have the opposite B here so they are symmetrical about the Y AIS so this is even all right question 39 to 42 the polinomial function case is even right so this guy even and the polom function p is out right question 39 find the values of A and B in the table above so we need to find a right here and we need to find B right here so we know that because the function K is even so we have K of s equals K of s great so we look at this one right here this guy and this guy right here so we have K of -3 equal 8 and then K of three right here must be the same thing equals eight right so the answer for eight here equal eight now look at this one because the function B is OD so we have of s equals P of s right so we have P of b equals three right here and then we have P of B E3 right so we look at this one right here right negative B SO3 is going to be right here which mean we have4 here so we just equate this so we have B = -4 and then B equals 4 so the answer is four now question number 40 K of 2 plus P of -2 K of 2 is going to be the same thing as K of -2 because the function is even right P of2 the same thing as P of two cor Because by the formula the function is even we have F ofx equal f of x if the function is OD f ofx is f ofx right so we just substitute values -2 K of -2 = 6 so could be6 minus p of two is going to be seven so Min - 7 so the answer is -3 now question number 41 K of1 is the same thing as K of one so K of one is going to be four now P of three is is the same thing as negative P of -3 right so it's going to be negative P of -3 is going to be two so it could be2 and the last one find the average of chance of P of s over the interval 5 to 5 right so we have average rate of CH is p of 5 minus P of 5 over 5 - -5 now now we look at this table we don't have P of five but we know because the function p is out so P of five is the same thing as P of5 right so we just plug in over here so P of5 minus P of5 right over 10 right so NE so p of5 is going be four so it's going to be -4 - 4/ 10 so it's going to be -4 over five right of the polom function f is shown above where the function is defined between 6 to six so we have the end points right here use the graph of the function f to answer the following now F has a local minimum add right local minimum is going to be this guy right here and this guy right here so at s equals -4 and three right so4 and three F has a local maximum at s equals Right Here Local maximum s to zero this guy right here6 and this guy right here is going be sit right the absolute maximum of f is the highest Val right here so the highest value right here is five at s = to sit the absolute minimum of f is the lowest one right here so it's going to be5 and S = to4 so it's could be4 right here now look at the second one the function G has local minimum as this guys right here so it's going be as s = to 4 and this guy right here s go3 the function G has a local maximum at this guy right here S5 and then this guy right here is could be zero you cannot count this one because you see the arrow which mean the graph keeps going up right the absolute maximum of the function CH is none because this guy keep going up you don't have the highest point right the absolute minimum of the function CH is this guy right here right it's going to be -3 at x = 4 all right number three I'm going to leave it for you this similar number four the graph of the polinomial function K is shown above which of the following could be an expression for K right so we need to know that this guy and this guy have the same n Behavior right so which means the degree must be even right so we know that this guy even now double pointing up so which means the leading coefficient is positive right so leading coefficient is positive right and we look at the zero right here we have one Z two Z and three zero but this zero right here because the function P off the xaxis so this guy multiplicity of two all right so we look at this one right here we see that the leading Co leading coefficient is negative we cross this one now right now this guy right here we look at question d right here we see that we look at the answer d right here we see that the power right here power one power one and power one right here so give be OD power so this WR right we need even power so we only have this one this one and we just check the uh zero right so the zero right here4 right so in order to write the polinomial it should be as minus the zero so which mean means minus -4 right and right at this zero right here we have multiplicity of two so this one must B squ right so negative negative turns out to be positive the answer should be B now look at this one right here the graph of the Pol function p is shown right here which of the following could be the expression for p right so we look at this one this guy and this guy have the opposite end behavior which means the function is are right and because the left hand side right here pointing up the right hand side pointing down the leading coefficient is negative right keep that in mind so we look at this one we see that leading coefficient is positive leading coefficient is positive so this wrong right and we have three zero right here 1 2 three all right so the degree must be three three right and we look at this one degree is four right this is two this is one this is one so degree four this is wrong so the answer must be C and we check this the zero right here3 so which mean s- -3 times the second zero right here s minus 0 right and the third zero right here s minus 2 and if we simplify it's going to be right here all right let's move on to question number six right here all right we look at this one this guy and this guy have the same end behavior right so the function is even and then because they both pointing up which mean the leading coefficient is positive right if the point in down the leading coefficient is negative right so we look at this one right here we see that the function has two real zero right here and they both of the sis right so so we have S minus -2 right and they both B of the x-axis so this guy must have the multiplicity of two so this guy multip of two and then this guy s - 3^ of two right and we know that the final answer must be even degree and then of course because the function has the imaginary root right so we look at this one right here leading coefficient is positive we need to cross out this one it should be even degree right so this one right here right at least we have four right here so we cross out a right now we look at B and D right here so you need to be careful because the function has imaginary Roots right here so the answer must be D now question number seven we need to find X where the function G of X greater than zero right so the function great than zero which mean is above the x-axis right so it's could be this guy right here all right so we put negative Infinity right here positive Infinity positive infinitive and negative Infinity right here so which means we need to find the domain of this one so could be negative infin 2 three right but because we don't have the equal side which means you don't include the zero so we have the zero right here as -2 we need to exclude this one so it's going to be negative Infinity 2 -2 Union -2 2 3 right so you need to be careful now question 7 b g of s less than equal to Z which mean means the guys below the AIS right here all right so it's going to be from 3 2 infinitive right now because we have the equal side which means we're going to include the Z so the Z is three so we have the bracket right here and we look at this one we need to include the zero right here too so this guy Union -2 now look at question eight right here all right question eight I'm going to leave it for you to practice all right question n we need to solve the inequality right here so the first step you need to do is to find the zero of this one so x + 4 = Z and then s = -4 now x - 3 = 0 and then S = 3 the last one is + 1 = 0 and then s = to1 now you need to draw the number line and then you need to place all the zero on the number line so I have -4 4 -1 and three the zero must be in increasing order so keep that in mind the left hand side right here you put negative infinitive the right hand side you put positive infinitive now you're going to pick a number within any interval and then you check to see if within that interval is positive or negative right so I look at this one right here from1 to three and my Pi is zero right so I subtitute over here so I have 0 + 4 0 - 3 0 + 1 right so I have four * -3 * 1 right and I look at this one it's going to be -12 and it's actually is negative right so within this interval right here I have negative right so I put a light right here put a light put a light right here right now if you see the multiplicity of each linear Factor right here as an OD number and then you keep changing the sign right this negative this guy positive this guy negative this guy negative this guy positive all right so finally we're going to look at this one right here we need this guy to be rated than equal to zero right so we only pick the positive one right here okay so the answer must be -4 to1 right and we have the equal side right here which means we need to include it with the practice Union 3 to infinitive write at negative infinity or positive Infinity you cannot put the bracket you must put parenthesis so keep that in mind all right so if you see the Ling coefficient right here negative number multiply both side by negative one right so I have three 3 * x + 2 * x -1 * x - 5 greater than zero you need to flip the inequality symbol right so then you need to find the zero of each of these linear Factor right here so I have Z of this one is going to be -2 this guy one and this guy five right so I play this guy on the number line right now I need to pick a zero within this interval right so I subtitute over here I have 3 * 2 * 1 * 5 which is positive right so this guy must be positive and then and then I look at this one I see that they all have multiplicity of OD number right so I keep changing the sign like this all right so I look at the second inequality right here don't look at the first one look at the second I need this one to be greater than equal to zero right so the answer must be this interval and this interval right here so because I don't have equal size so I don't need to include the end points right so it's going to be -2 to 1 Union 5 to Infinity all right now look at number 13 so we need to find the zero of this one so it could be S = to Z with multiplicity of two right right here and then s = to -7 for this one and then s = to four right so I draw the number line so I so I need to place -7 zero and four over here right I put a line right here the left hand side I put negative infinitive the right hand side I put positive infinitive right so I look at this one right here I need to pick a number I pick number one right here subtitute one into this inequality so I have -3 * 1 * 8 * -3 right so it's going to be positive right so this guy positive I put this a positive right now four you look at four right here this is multiplicity of one right so I keep changing the sign now zero right here multiplicity of two and then the side doesn't change this guy positive this guy must be positive right keep that in mind when the graph passing the zero with multiplicity even right the side does not change keep that in mind and then this guy right here7 this multiplicity of one so this guy change right negative now I need this guy to be r for so which mean my answer right here all right so it's going to be -7 comma four right and then I have the equal side right here so I need to include the end point all right now the last one right here I need to find the Z of this one so the zero of this one is going to be S equal to zero multiplicity of three so I don't care right keep in mind I only care about the multiplicity of even power right this guy s equal 8 multiplicity of two all right and S = 5 multiplicity of One S equal 1 multiplicity of two right so I draw a number line I place all this number on the number line so it could be8 go first and and then zero next one one right and then the last one should be fine right so I draw the line right here put positive Infinity right here negative Infinity right here so now I'm going to pick a number two within this interval and then I'm going to subtitute over here so I have 4 * x q which means four * 2 power 3 * 10 power 2 * 2 - 5 is going to be -3 * 2 - 1 is going be 1 so 1 square so this guy positive positive positive this negative right so this guy must be negative right so the answer is negative so this is negative now you need to be careful look at F here okay so this is the zero right X = to 5 this guy multiplicity of one right so I keep changing the sign X = to 1 multiplicity of two so the side doesn't change so it could be negative right here x to Z right here multiplicity of three so because this a odd multiplicity right so this guy right here keep changing the the sign right because multiplicity of three now 88 which mean the zero up this one right here multiplicity of two so this guy doesn't gen asde so this guy positive right now I need to conclude I need this one to be less than equal to zero so the answer must be this right here right so it's going to be 02 5 now I need to include the end point so it could be 0 to five right here and then right here you see this guy multiplicity of two right here right s = to8 right here this guy make the inequality equal to 038 so this going to be Union 28 all right let's move on to the next one number 15 right here we need to find all real zero and indicate the multiplicity of each zero right so the first one zero is three right because we let s - 3 = to0 S = to 3 which multiplicity of four right the second zero is going be -2 multiplicity of one right because this a power one this guy same power one so this could be one multiplicity of one right this guy right here the zero right here s right here equal to zero so the zero should be zero multiplicity of one the second one -3 multiplicity of one right this one and the thir one is going be two multiplicity of three right okay so this one right here this one right here I'm going to leave it for you as a practice right so I look at number 19 right here now this one you need to factor this one now right so s Square minus 9 you're going to use the formula a² - b² = a - b * A + B right so this one is going to be x - 3 * x + 3 right and then this one right here I use this square formula so give be a squ + 2 A+ B squ is going to be a + b all together square or you can factor trinomials and you still get the same thing so it's could be x + 3 all together Square right so I'm going to combine so s - 3 right and this guy x + 3 * x + 3 squ is going be S + 3 Q right so the Z right here the first Z is going be right here s - 3 so X = to 3 with the multiplicity of one right this is one and then the second one X = to -3 right here with the multiplicity of three correct all right the next one number 20 right here I need to use the different of two square to factor this one so it could be equal to -2 x * x + 2 * x - 2 right and this one right here it turn start to be complete Square so if I have a² - 2 a b + b² is going to be a - B all together squ right so this guy turn out to be S - 2 all together squ now this guy right here I use the S method to factor this one so I put C right here it's going to be8 and I put B right here it's going to be2 I need to find two integers that multiply equals 8 add up to -2 which is -4 and 2 so this guy turns out to be S -4 4 * x + 2 right so then I'm going combine I have -2 x right here now s + 2 s + 2 right here it turns out to be S + 2 squar right s - 2 s -2 right here I combine this s - 2 CU right and the last one should be s s -4 right so we have the first zero let's s what Zer right here this guy right here so s = to Z and this is multiplicity of one second one s = -2 multiplicity of 2 next s = to two this guy right here multiplicity of three next s = to for multiplicity of one right let's move on to number 21 right here so I just Factor this one out so it's going to be S time x² - 3 x - 10 and then * x² + 7 x + 10 right so the first one right here I use S meod right here so I put -10 right here -3 right here right I need to find two integer multiply -10 and add up to -3 which is5 and 2 so this one can be written as s Time s - 5 * x + 2 right now the second one right here I need to find two multiply equal to 10 add up to S right so it could be five and 2 right so it can be x + 5 * s + 2 right so I can rewrite this one as s s + 2 s+ 2 is going be S + 2 square right and then s - 5 * x + 5 right so I need to find the zero right so the first one is going be X = to Zero this one right here multiplicity of 1 s = to -2 for the second one right here multip it of 2 next s = to find multiplicity of one right and the last one s = to5 multiplicity of one right this a power one now let's move on to number 22 right here we're going to factor this one out so it could be S * s² + + x - 12 and this one factor s² out so it can be x² * x² - 9 right so you need to factor this one using S method like what I did right here it turns out to be S time x + 4 * x - 3 and this one right here S squ and S right here I combine it's going be S CU and this guy turn out to be S - 3 * x + three right okay so you need to find the Euro so the first one s = to Euro multiplicity of three the second one you're going to combine this guy and this guy right here so it could be square so I don't need to write this one so okay be S = to -4 multiplicity of 1 S = three multiplicity of Two And The Last One S equal to 3 multiplicity of one right this power one this power one okay so the last 23 and 24 I leave is for you to practice now let's move on to question number 25 so given limit of the function f of x when s approaches infin equals infin and limit after the function f s when s approaches positive Infinity equal negative Infinity so the polinomial function f has and behavior above so which one could be right so we look at this one we see that limit of the function f of x when X approaches negative Infinity equals negative Infinity so we need to SC the r graph right here all right so we see that this one is y this one is s we need to put positive infinitive play here this guy negative infinitive this guy negative infinitive and this guy positive Infinity right so we see that when X approaches negative Infinity which mean X go this way the function approaches negative Infinity fun the function going down so which mean the graph should be something like this right now when s approaches positive Infinity the function approach is negative Infinity right like this so the end behavior should be something like this so this guy this guy and this guy have the same end behavior right so which mean we know that this is even degree right and then we see that they they both pointing down which mean the leading coefficient is negative right negative so we look at this one even powers so we need to cross out b and d and leading coefficient is negative so we're going to pick C the answer a is wrong right now number 26 we do exactly same thing right so we scale this one so this is s this is y so we're going to put positive infinitive negative infinitive positive infinitive negative infinitive right so limit of the function G of X when X approach is negative Infinity is positive Infinity right so X approach is negative infinitive right the function approach is positive infinitive is going be like this all right so the end behavior should be something like this right now when s approach is positive infinitive this way the function approach is negative infinitive like this so the other end Behavior should be like this all right so we see that they have opposite end behavior which means it should be a degree right okay and then because the function pointing up to the left which means the leading coefficient is negative all right so we look at this one right here not degree we need to cross out this guy and this guy leading confusion is negative we need to pick D now number 27 we need to scale the graph right so this is X this is y and then we need to put positive infinitive right here positive infinitive negative infinitive negative infin right now when s approaches negative Infinity S go this way the function approaches this way right so we have n Behavior should be like this and S approaches positive Infinity the function approaches positive infinitive so the other end behavior should be like this so they have opposite end behavior which mean this are decree right and then because the left hand side right here pointing down which mean the leading coefficient is positive right so based on this information we leading coefficient is positive so we need to eliminate c and a right because we have -2 and -2 right here now we check the power this is five this is two and this is one add up to eight so this is wrong right and the last last one right here five 2 and two right here is going to be nine and leading coefficient is positive two so the answer must be D all right the last one we have last case of s equal this one and which of the following pass our statement about about the n behavor of K is correct right so we look at this one right here the highest degree right here so we see that the highest degree is going to be four right so this see even and then the leading coefficient is negative correct so we know that the degree is even so they both pointing up or pointing down like this okay and then because the leading coefficient is negative right leading coefficient is negative so we're going to pick this one right here so which mean when s approaches negative Infinity the function approaches negative Infinity when s approach is positive Infinity the function approach is negative Infinity right so s approach is negative Infinity function approach is negative infity so the answer is a now let's move on to number 29 given the table right here which of the following claim and explanation statement best fix the data great so question a f is best model by a linear function because the rate of chance of consecutive equalent input value intervals is a constant we're going to check this one right here we see that this one add two each time right and the right hand side right here negative in order to get negative 1 from -2 we're going to add one right and this one we're going to add four right so this is not linear right because 1/ by 2 is going to be 1 12 4/ by 2 is going to be two the average R of chain is different it's not a constant so this wrong now question b f is best model linear function because the rate of CH of our consecutive equalent input values interval is a linear right this one at 7 right so 7 IDE by 2 is going be 7/ 2 equ question a is WR question B must be wrong right because the function is B model by linear function the rate of change over consecutive equalent input value intervals must be constant right so this is not a linear so this is wrong now question c f is best model by quadratic function because the rate of chain of a consecutive equalent input value interval is a constant no this is not right right this must be linear right yes this is right so we're going to check this one right here we see that 4/ two is could be 4/ two right this is 7 / two and this is 10 / two right now we need to find the chain of this one right so this one right here in order to get 4 over two from 1/2 over from 1/2 I'm going to add three and half right this one I add three and2 this one I'm going to add three and2 right so this one tells us the answer must be D we look at this one right here we see that every single times we're going to add 10 right okay the right hand side right here this guy decreasing by 40 this guy decreasing by 20 this guy decreasing by 10 and this guy in decreasing by five right so I need to find the rate of change so this one the re CH is -40 / by 10 this guy - 20/ 10 this guy 10/ 10 and this guy5 / by 10 and going simplify this one is going to be -4 -2 1 and one2 and you see that average rate of chain is increasing right4 2 -2 2 1 and 1/2 which is increasing so because the a rate change is increasing the function could be concave up right so the answer must be B now let's move on to the next question the graph of the function f is shown above which of the following statement about the function f is correct so the function f is positive and increasing no this not a positive is only positive in this interval only right so this wrong the rate of chain of the function f is positive and increasing yes you can draw the tensent line you see that the Christmas tree say if the tangent L is going this way the slope is positive and if the slope is positive the rate of change positive right and if the rate of chain is positive the function is increasing right so this is true now the function is increasing so the answer must be B okay question number 32 the gra function G is which of the following statement is true about the function G question a the function G is an even function yes because they have the same end behavior and the rate of change of ches is increasing now we need to draw the tangent line right here the first tangent line the second tangent line and the third tangent line right here you see that if the tangent line is going this way the slope is positive right this guy zero this guy is negative right so from left to right the the rate of change of the function G is decreasing right so this is wrong question B yes that's right the rate of change is decreasing the function is even now question 33 the the function H is increasing we draw a tensent line yes this is the slope is positive the function is increasing yes and the rate of chain of H is increasing we just draw another tension line we see that now tent light right here and you see that the tangent line getting less step which means the slope is smaller rate of chance of the function H is decreasing so the answer a is not right the answer must be B right here decreasing the function is increasing now question 34 right here the function K is increasing we draw the tangent line right here and you see that the function the line is going this way the slope is negative which means the function is decreasing so we need to eliminate a b right now we draw another tangent line right here we see that the tension line getting lasted and then this go in this way right here right the tangent line is steeper the slope is smaller right so which mean the slope of this guy right here is less than this guy right here which means the function has an in increasing rate so the answer must be C now let's move on to the next one so we are at the home screen we need to add graph right here and we input the function it's going be negative plus 30 3.1 okay so you hit enter all right so it's going to be like this we need now we need to restrict the domain from -3 to 3 all right so we need to hit [Music] menu window Zoom right here and then we're going to go to window setting right here s mean we put -3 and S Mas we put three right and y means y mask we just leave it so we hit enter so it's going be like this the function f has a relative minimum so we have two relative minimum the first one is going be right here and the second one is going be right here right so relative minimum right here we just hit menu analyze graph and minimum right here okay wait move the line to the left of minimum hit enter move the the line to the right up minimum hit enter so it could be 0.718 so it could be0 718 so keep that in mind the question asks you to find the relative minimum s the value of s right here and the right hand side right here we we know that this is the lowest point right here it's it's going to be at three right here so okay now relative maximum we do exactly same thing we hit menu analyze graph maximum move the line to the left up maximum move the line to the right so it can be 1.99 right here so it can be 1.99 and the right the left one right here is going be at s = to-3 right so -3 right here on what intervals is the function is increasing right the function is increasing from the interval 0.718 to 1.99 right this one right here the middle one right here because the tangent Lin has the positive slope so we have the interval netive 0.718 to 1.9 n right now on what intervals is the function f is decreasing which means the other one right here it's could be-3 to this one and it's could be from one 99 2 3 right so it could be [Music] -320 718 Union 1.99 2 3 right right here you can include the N Point right here3 and the right hand side should be three right now question two and question three you going to do exactly same thing all right so I just leave it as your practice now let's move on to the last one right here number one which of the following statement about the Euro of the function f is correct right now look at this one we have one Z right we have two imaginary zero right here s² + 4 S s + 4 = to0 right so it's going to be X = to x² = -4 so we have two imaginary zero or two non real zero right so we look at this one right here we see that two non real zero so we need to eliminate A and B right now we look at this one right here we have the zero be S = to Zer this one right here we have the z s = to 3 this one we have S = to 1 so we have three distinct real zero and two non real zero so the answer must be C on to question two given the function G of x on what intervals is G of X greater than equal to zero right so we need to factor this one out right so we have a * c = 2 * -12 is going to be -24 and b equal to 5 now we need to find two integer multiply equals 24 add up to five it's going to be 8 and -3 all right so I have x + 8 * x - 3 now I look at this one I see that the leading coefficient is two so I need to divide this guy by two this guy by two I simplify aide by two is going be four so I have S + 4 * this one I cannot divide 3 by 2 so I need to multiply this one so it could be 2 x - 3 now I need to find the zos of this one so the Z should be -4 and 3 and2 right so I put this guy on the number line it's going to be4 here 3 and half right here the left hand side is negative Infinity the right hand side is positive infinitive I put the L right here all right so I pick a number within this interval so I pick zero I subtitute over here right so I have 0 + 4 is going to be 4 0 - 3 is going be 3 4 * -3 is going be negative So within this interval it's going be negative so I look at this one I don't have the even multiplicity of the linear Factor right here so I keep changing the size so this guys will be positive this guy should be positive now I look at this one right here I need this guy to be greater than equal to zero So my answer should be this interval and this interval so it's going to be negative Infinity 2 four and then Union 3 and half to infinitive right right here I put the pothesis but right here I put the pret because this guy right here both zero up the inequality right here and this guy right here we need R than equal to zero right so the answer must be D now question three the polinomial function has digital three multiplicity of two4 multiplicity of three this guy this guy what is the least possible degree of this one all right so we we need to start from here right the function h of s has the zero S = 3 all right x = 3 and then we subtract three both sides okay you we have the linear Factor s - 3 = to zero so s - 3 right power of two because the multiplicity of two now next One S - 4 multiplicity of three so I have S = to -4 add for both sides s + 4 = to zero so I have x + 4 right here multiplicity of three so I put three right here power three now I have the next One S = 2 I this is imaginary number so whenever you have imaginary Ru it conjugate must be the second root so keep that in mind if you have A+ b i is a root and it conjugate is a minus B I must be the second root so we have S - 2 I right and then the other one should be S + 2 I right and the last one right here this is same thing s - 5 - 3 I and S - 5 + 3 I all right so now we're going to check the degree right this is power two power three this is power 1 power one power one power one so it's could be power 9 so the answer must be D question number four the graph of the function H is shown above which of the following statement about the function H is correct right so we look at the rate of chain of H is positive which draw the T line right here right and we see that the tensent lines going this way which mean the slope is negative right this guy slope is positive so because the slope is negative the rate of chain is negative right so we need to eliminate the answer a right answer B the rate of change is negative which is right and decreasing we draw another tangent line right we draw another tangent line right here and if it's going this way if the tangent line getting steeper the slope is smaller right which mean it's decreasing right so the answer must be B now question number five the function case is given which up the following statement about the zero and multip lity of K is correct right so we just Factor this one it's going be S - 3 time x + 3 this one already linear Factor so s + 3 q squ and this guy turn out to be S + 3 squ right here right so we just combine them it's going to be S - 3 * x + 3 so we see that 1 + 2 + 2 is could be five right so we look at this one the zero must be S = to 3 S = to -3 multiplicity of five so the answer must be a now look at number six given this table which up the following statements pass about the function G could be true okay the function G's is an odd function and the graph of the function G's is concave up so we need to find the aage rate of change right so this one right here we need to add two add two add two and add two the right hand side right here we need to add one add four at five at sit right okay so we need to find the average rate change of each interval right so the first one is going to be 1 and2 the second one is going to be 4 and2 the third one is going be 5 and2 and the last one is going be sit and half right and we see that the average rate of change is increasing right so because the average rate of chain is increasing the function is concave up so because the function is concave up we need to eliminate A and C right now next one in order to know if the function is not even we apply the formula f of s is f of s this is out F of s is f of s this is even right this guy right here so I have G of -1 equal -3 right now I have G of negative of negative 1 which means negative of look negative s right here NE 1 is G of one right but G of 1 equal to three right and this one equals negative G of 1 right so I just prove that g of 1 equals g of1 right which is the formula right here so the function is OD the answer must be B now let's move on to the next question question number seven right given the function is even right so if the function is even you have F of s equals F of s right so which means s opposite right s opposite and then the output must be the same right same output but opposite s value right so we look at this one right here a we look at a right same output now look at this is two and then we need opposite s value right here right same output so a equal to one right now b b right here and then I see that the output is five and look at the output output right here five right so B right here must be opposite to -3 so which means B = to 3 right now C right here C and I look at the input here sit and then input must be opposite right so sit and negative sit right here so C must be equal to4 so4 so if you add them up the answer must be zero so the answer is C all right look at number eight the function which out the following statement about the function G is correct the function G is increasing at an increasing rate or something like that okay so you just draw a tangent line right here and you use the Christmas tree trick right the tensent line is going this way the slope is positive the function is increasing the tangent line is going this way the slope is negative the function is decreasing right so because the tensent line is going this way which mean the function is decreasing so I eliminate A and B all right now I need to draw the second tension line right here and I compare the slope of this guy and this guy all right so because this guy right here has smaller slope of this guy because this gu is steeper than this guy so if this guy has a smaller slope of this guy which mean the the slope is increasing right so the function is decreasing as an increasing rate the answer must be C I'm going to leave question 9 12 11 12 for you to practice let's move on to question number 13 let the function f be an odd function such that F of -4 = to 5 is the location of local minimum which of the following statements must be true all right so because the function is out all right we have F of s equals F ofx so which means the input opposite and the output is still upis right because they both symmetric about the origin right so we look at this one this is4 this turn out to be four this five turn out to be negative five right so I have F of 4 = -5 right and this is local minimum this guy becomes local maximum so the answer must be B all right so I'm going to leave question number 14 15 for you to practice right this is similar to the previous one now question number 16 we need to write the polom function of this one right here so we need to find the zero right here the zero right here the zero right here right and we see that this guy House of the x axis so this guy right here has the multiplicity of two right this guy has the multiplicity of two and this guy multiplicity of one so total must be degree F right so I look at this one right here I see that this is degree one degree one so this is wrong degree one one one this is three so wrong this is 2 2 one so this is could be right this guy 22 one this is could be right right now I look at the end behavior right here this guy and this guy opposite end behavior which means the leading coefficient is negative so the answer must be C all right now number 17 all the way to number 24 I'm going leave it for you to practice and and it's good time to stop thank you for watching I hope you enjoy the video remember that if you failing to prepare you repairing to fail like T buildings as the chemicals it take us higher the night's young it it's just be good and she puts her hand in my