hi everybody I'm Judy from the AP program welcome to the AP pre-calculus prerequisite review series our goal for this series of videos is to help you feel confident that you have the skills and content knowledge necessary to be successful in AP pre-calculus if you've taken Algebra 1 geometry in Algebra 2 then you've got these skills and you're ready for this class so algebraic manipulation of linear functions is our first review topic let's get started with each prerequisite topic we'll start with an instructional video just like this one then you'll have the option to complete a practice problem set check your answers using an annotated Solutions page and watch a video that shows the problems being worked out if you had any difficulty with any particular problems when you complete the components for each lesson you'll be able to decide if you need extra help from a teacher or a tutor or a peer and if you do need some extra help or support you'll know exactly what type of problems you need that support with so that your time together can really be concentrated on the things that are going to help you most in the end when you're done with the 13 sets of videos practice problems and solution sets you'll be able to be very confident that you are ready for Success going into this AP pre-calculus course and then you have all of the skills necessary to do well in the course when you have the right math skills applying them in a new situation makes it easier to understand new content and you've developed a strong Foundation of prerequisite skills and content knowledge in Algebra 1 geometry and algebra 2. on the screen you'll see the mathematical skills that you'll master in AP pre-calculus and we'll come back to this at the end of this lesson and again at the end of each other lesson or instructional video so that you have a chance to think about the skills that you already have and the skills that you'll be building on in AP pre-calculus next year and that's going to help build your confidence remember this is a video and you can pause it at any time to take notes and when we get to examples you'll be able to test your knowledge by pausing the video and solving the problem on your own before you see the solution this is going to give you a good sense of whether or not you may need some extra practice with the topic so let's get started do you remember the three forms for a linear equation the first one is slope intercept form and that looks like f of x equals MX plus b where m is the slope of a line the slope being the change in y over the change in X or another way to say it would be rise over run that should be a really familiar topic and a really familiar statement to you because you've been working with slope intercept form equations since you were in algebra 1. these are equations that are best to use when you want to graph a line the second form is point-slope form and that's y minus y1 equals m times the quantity of x minus X1 again m is the slope just like it was before and X1 y1 is a point that is on the line it can be any point on the line point-slope form is best to use if you have to determine the equation of a line and you know some of the points on the line already the third form equation is a standard form equation and that's ax plus b y equals c where a B and C are constant values the x-intercept or the point where the line crosses the x-axis is C over a zero and the y-intercept or the point where the line crosses the y-axis is 0 C over B this is a form that's often most efficient to use when you're solving a system of equations and then you have the the items that are true for all forms of linear equations which is the rate of change is constant the domain is going to be x equals all real numbers unless it's limited by some sort of a real life application and the exception to that rule is that this the domain would be undefined for a vertical line the range or the Y values that are possible would again be all real numbers unless it's limited by some sort of real life situation and the exception again is when your line is a constant vertical line sorry constant horizontal line all right so here we are with example one a line passes through the points negative two negative 4 and negative 5 5. find an equation for the line in all three forms of linear equations so your first thing to do is to think about what questions do you need to ask yourself first and foremost you need to ask yourself which form of equation should I start with and why and then once you decide on a form of equation you should be asking yourself which what values do I need to calculate or find and and how do I go about doing that so because we are given two points the most efficient form of an equation to start with is point-slope form once we have our point-slope form equation it'll be very easy for us to derive the slope intercept form equation and the standard form equation if I'm going to write an equation in point slope form I have to know the slope so that's where I'm going to start to find slope of an equation using two points you find the difference in the Y values and the difference in the X values and you write the ratio for that so we're going to start by subtracting 5 minus negative four the difference in the Y values and negative 5 minus negative 2 the difference in the X values when I do that I get 9 over negative 3 which simplifies to negative 3. so the slope of this equation is negative three that's going to replace the m in point-slope form to replace the X1 and y1 in point-slope form I just need to decide which point I'm going to use it does not matter which one I use either either one of the coordinates is fine it will end up giving me the same slope intercept form equation and or the same standard form equation regardless of which point I choose to move forward though I'm going to go ahead and make a decision I'm going to use that first set of coordinates the negative 2 negative 4. I'm going to replace X1 with negative 2 I'm going to replace y1 with negative 4 and I'm going to replace the slope or M with negative 3. when I do that the point-slope form equation is going to read y minus negative 4 equals negative 3 times the quantity of x minus negative 2. this can be further simplified we don't leave things with minus negative instead we change it to plus signs so it turns into y plus 4 equals negative 3 times the quantity of X Plus 2. that's my point-slope form equation my next step is to change this into slope intercept form slope intercept form y equals MX plus b that means I need to get y all by itself and that means I can't have any parentheses so I'm going to do two different steps I'm going to subtract 4 from both sides of the equation I'm also going to go ahead and distribute this negative 3 negative 3 times x and negative 3 times 2. when I do all of those steps I subtract 4 from both sides of the equation I distribute my 3 so that I have negative 3 times x and negative 3 times 2. when I simplify that I get y equals negative 3x minus 6 minus 4 and I'm going to simplify that a little bit more to say y equals negative 3x minus 10. so there's my slope intercept form equation next I'm going to turn that into standard form a standard form equation is ax plus b y equals c so you heard me say X and Y on the same side of the equal sign which means I need to take this negative 3x and move it over so I'm going to add 3x on this side of the equation add 3x on this side of the equation to get to my standard form and lo and behold my standard form is 3x plus y equals negative 10. so in our first example we took the time to create equations of a line but in this example we're actually going to sketch the graph of a line based on some of the things that we know about the form of the equation that it's in when we look at this example and we're asked to sketch a graph of the line Y equals negative 2x plus 5. the questions that you should be asking yourself at the start are things like which form of equation is this and what does that mean to me what numbers do I need to have on my x-axis and my y-axis and where do I begin my graph and how do I find additional points so this particular equation is in slope intercept form that's going to tell us all the things that we need to know in order to graph the line we're going to start by identifying the y-intercept in the slope intercept form y equals MX plus b b is the y-intercept so in this case it's 5 or the point 0 5 that's where the line is going to cross the y-axis the second thing we'll do is identify the slope of the line the slope of this line is in place of the m in y equals MX plus b so the slope of this line is negative 2. negative 2 can be written as the fraction negative two over one and that can be interpreted as y decreases by 2 when X increases by 1 or you can interpret it the other way y increases by 2 when X decreases by 1. so now we're going to go ahead and sketch the graph when we sketch the graph we're going to start at the point that we know which is the y-intercept so we'll start at the point 0 5. then we'll use the slope to find additional points we'll start at 0 5 and from there we'll go down 2 and to the right one to find additional points or we'll go up 2 and to the left one to find additional points the result is going to be the graph that you see here on the right you can see that it crosses the y-axis at positive five and if you go down two into the right one you can see that another point on this line is 1 3. and if you go down 2 and to the right one again you can see that another point on this line is 2 1. so on and so forth so when you have a line in slope intercept form you you graph that equation by starting at the y-intercept which is a point on the y-axis and you use the slope to find the additional points so that you can draw your line all right so moving on to example three you're going to be asked to sketch a graph of the line 3x plus 2y equals 12. and first we're going to start with what questions should we ask ourselves and again they're very similar to the last time first question is which form of equation is this and what does that mean as far as graphing something is concerned the second question is what numbers do I need to have on my x-axis and y-axis and the third question is what's the easiest way to go about finding points that fall on this line well let's start at the beginning this is a standard form equation and if you remember back on the very first slide that we started reviewing on I set a standard form equation would help us very easily graph a line right so this is going to be an easy thing to do step one is going to be identifying the y-intercept to find the y-intercept I need to think about what happens when I put 0 in place of x and when I do that I get the equation 2y equals 12. so that means the y-intercept is 6 or the point zero six now I'm going to do the same thing with the x-intercept in order to find the x-intercept I put a 0 in place of y when I put a 0 in place of Y I get the equation 3x equals 12. that means the x-intercept is 4 or it falls at the point 4 0. so I know that my y-intercept is 0 6 I know that my x-intercept is 4 0 I'm ready to sketch the graph I'm going to put a point on the y-intercept at 0 6 I'm going to put a point at on the Y on the x-intercept at 4 0 and then I'm going to sketch the line so that's what you see in this particular graph we've got the y-intercept at six we've got the x-intercept at 4 and we've got the line that goes through those two points all right moving on to something a little bit more complex because when kids see words they Panic automatically but don't panic because this is actually pretty easy for example four we're dealing with a car rental company who charges a base fee plus a constant fee per mile driven the table at the right compares the total distance driven in miles with the total cost of the rental in dollars and you're asked to calculate three things first what is the per mile fee second what is the base fee before any miles are driven and third write the linear equation that models this situation and identify the domain and range so we're going to take note that if you drive 110 miles it's going to cost you 186 dollars if you drive 145 miles it's going to cost you 207 dollars and if you drive 170 miles it's going to cost you 222 dollars the per mile fee can be calculated in a different in a variety of different ways you can compare what happens to the cost when you go from 110 to 145 miles or compare the cost from 145 miles to 170 miles or compare the cost from 110 miles to 170 miles so here's what happens if we look at 110 to 170 we've driven a total of 60 miles and it's cost 36 extra dollars if we go from 110 to 145 miles we've driven a total of 35 miles and it was a total of 21 extra dollars and if we go from 145 to 170 miles that's a total of 25 miles driven for a total extra cost of 15 dollars regardless of which one of these you use your total cost is 60 cents for every mile so 36 dollars divided by 60 miles 21 divided by 35 miles or 15 divided by 25 miles each of those is going to come out to exactly 60 60 cents per mile the second thing that we're asked to find is the base fee the base fee is what you would get charged before you start actually driving the car right that's the amount that you that you're charged just for getting the keys to the car and getting in it to drive it off the lot so that base fee can be calculated if you just find the cost of the miles for any of the distances driven and you take that away from the total so if I find the cost for driving 110 miles at 60 cents a mile and I take that away from the total cost of the rental that'll give me my base fee driving 110 miles at 60 cents per mile would be a total of 66 dollars if I take 66 dollars away from 186 dollars I'm going to find out that my base fee is 120 dollars it would have come out the exact same if I had done the same math with the 145 miles and 207 dollars it would have come out the exact same if I'd done the same math with 170 miles for 222 dollars the base fee would calculate to 120 dollars regardless so now the third part of this write the linear equation to represent this situation 60 cents per mile is the slope of this linear situation because it's happening over and over and over again is constant that you add on 60 cents for every Mile right so that's your slope that's your m the base fee is the value that you start with the base fee is your y-intercept so if you substitute those values into slope intercept form y equals the price per mile which is 60 cents times X plus the base fee which is 120 dollars your slope intercept form equation would be y equals 0.6 X Plus 120. so nothing too strenuous here right basically a little bit of math a little bit of math and then putting numbers into the equation that we knew that it would have to go into so the domain for this particular situation is also something I was asked to find and the domain would have to be positive numbers right if I was driving no miles at all I wouldn't even be renting a car and if I was driving and it's not possible to drive negative miles so the domain has to be X is greater than zero and the number of miles would always be rounded up to the nearest whole number and then for the range the range in this particular case starts out at 120 dollars except that again it's not going to equal 120 dollars because I'm not going to rent a car and drive it nowhere so it will be Y is greater than 120 dollars and that's example four so now with example five let's try something sort of similar to what we just did in the last in the last um last example the water is drained out of a bathtub the liters of water that are left in the bathtub as a function of time is shown in the graph to the right so you've got the amount of water in liters that are in the bathtub you have the amount of time that has passed in minutes right and you've got the line that represents the water compared to the time we have three questions how long did it take to completely drain the bathtub how much water was in the bathtub when it started to drain and then as we did before write a linear equation to model this situation and identify the domain and range so part A how long did it take to completely drain the bathtub that's when the water level is at zero and you you can see here that that happened at five minutes the second part how much water was in the bathtub when it started to drain that's before any time passes right so it starts out at 360 liters of water and then in order to write the equation we're going to calculate the slope and we've already identified the y-intercept to calculate the slope we've got the point five zero and the point zero 360 to work from so if you subtract the Y coordinates 0 minus 360 is negative 360. if you subtract the x coordinates 5 minus 0 is 5. when you divide those you find out that the slope is negative 72 liters per minute the water is Flowing out of this bathtub at 72 liters per minute the y-intercept is where the line crosses the y-axis which is at 360 liters and so the slope-intercept form equation is y equals negative 72x plus 360. so really this is all about just interpreting the graph right and then determining what our slope is and writing the equation pretty simple in terms of domain that is between 0 minutes and five minutes the water is still in the bathtub from zero minutes to five minutes and that's inclusive right we start out with no time passing we pull the plug and at five minutes or when it equals five minutes the water is completely drained and then the range for liters it starts out at a maximum of 360 liters it drains down to no liters and it can equal 360 and it can equal zero so again the range is between 0 and 360 including those end points so here we are back at the skills that you'll use throughout AP pre-calculus the problems we looked at today demonstrated a foundation for skills within all three of these AP pre-calculus mathematical practices and I hope that you see that that's true so that you can have the confidence that you already have the foundation for pre-calc hopefully all of this helped remind you that you are a Master with linear equations and you're gaining confidence for your success in AP pre-calculus thanks for joining me I hope you'll be back for the next lesson you've got this