Algebra study guide - Algebra 1 final exam review video 60 fully worked out problems to help students prepare for an end of course exam Prepare for algebra 2 or college algebra Topics covered Try problems on your own System of equations problem Solution to a system of equations problem Solving a system of equations by graphing Solve a system of equations by the substitution method how you solve a problem involving the substitution Solve this system of equation by elimination or addition method Evaluate an expression Exponents Product Rule Area of a rectangle involving polynomials Exponents Power Rule Exponents Quotient Rule Volume of a cube involving polynomials Exponents Quotient Rule Find the degree of a polynomial Find the height of an object after t seconds using a polynomial Combining like terms in polynomials Polynomial addition problem Polynomial subtraction problem Perimeter of a triangle using polynomials Polynomial multiplication problem Multiply binomials using FOIL Multiplying polynomials Add binomials and multiply binomials (polynomial addition and multiplication) Binomial squared problem Difference of two squares formula a^ 2 - b^ 2 = (a + b)(a - b) Find the GCF power of the variable and the variable has to be in every term for it to be in the greatest common factor of all three terms so we have a 12 an 8 and a 9 so we go with the lowest so x to the eth and then we have a y to the 4th y 5th and y to the 6 so the lowest y 4th so this is the complete GCF x to the 8th * y 4th this next problem is asking us to factor out the GCF we can see that in our coefficients 14 12 and six that they're all even numbers so we have a factor of two so our GCF we can at least pull out a two and for the x's we always go for the lowest power so x or x to the 1st so that's what we will be factoring out in front and we've got to figure out what goes here it'll be three terms so 2x * what will give us 14 x cub 2 * 7 is 14 and x * x^2 is x cubed for the next term 2 * what will be 12 and that's 6 and x * another x would give us x^2 for the last term 2 * 3 would give us six and we already have the x so this is considered the fully factored answer this next problem wants us to factor this tromial so we want to see if we can factor it into two binomials since the first term is x^2 we know that these two will be x * x and for the two numbers that go inside these binomials we're looking for two numbers that multiply to equal 10 that will also add up to -7 what multiplies to be 10 we've got 1* 10 and 2 * 5 but to add up to a negative we're going to need negatives on these and we could see that this is the pair -2 + -5 that would add up to -7 you can always check these problems by foiling them or multiplying them through so if we distributed like the x x^2 x * so this is the check x * -5 is -5x and -2 * x is -2x -2 * -5 is pos10 and these add up to -7x so we can see that we get back to the original problem which tells us our factoring is correct as long as there's no GCF remaining inside these two parentheses and there's not number 28 is asking us to factor this tromial there's no GCF um involved because um the first two terms have an x in them and the last two have y's in them but none of them have an x and a y and as far as the coefficients the lowest one is a one which we don't factor out a one so we are going to check to see if this factors into two binomials again this is x^2 so it'll be x * x this last term has a y squared on it so we know that these two have to have y's so now we just need the numbers in front of the y's so we're looking for two numbers that multiply to be -3 that will add up to -12 so since 13 is prime the only way to get -3 is either 13 * -1 or -3 * 1 the pair that adds up to -12 is -3 and one so one of these is minus3 and the other one is + one but we don't need the one in front of the y again you can check these by multiplying it through to see if you have the correct answer so x * x is x^2 x * y is xy and -3 y * x is -3xy and -3 y * y is -3 y^2 these two middle terms will add up to -12x y which will get us back to the original problem so we know our factored answer here is correct for problem 29 they are giving us one of the two factors and asking us to factor the problem basically they just want us to fill in this binomial so these are a little bit easier to do than if they hadn't given us the 10 x - 3 because we could just check these two things 10 x * what would give us 50x^2 and 10 * 5 and x * x would give us the 50x^2 when we multiply those two for the last two terms -3 * pos1 would give us -3 again we can check to make sure that this works 10 * 5 is 50 x * x^2 10 x * 1 is 10 x -3 * 5 is -15x and -3 * 1 is -3 which I'll just write as -3 so 50x^2 - 5x - 3 does check number 30 is asking us to factor this tromial it's the harder kind of tromial where a does not equal one because a is two here so because of that we know that one of the factors has a 2x and the other one has an x in order to multiply to be 2x^2 we learned a few different methods for factoring these harder type tromials we could use grouping or we could use guess and check we could also use slide and divide for this problem since two is prime and 13's prime there's only going to be four options if we use guess and check so I'm going to go ahead and use guess and check so by guess and check you just fill it in and see what works right so here's three of the four options and here would be the fourth and let's fill them all in and what do I mean by that well these two numbers have to multiply to be -13 but they're each multiplying off different things either the x or the 2x we can either have -1 * pos3 like that or if we reverse the order we could have positive3 and -1 like that they're multiplying off of different terms here so those would be two different answers we could also change the sign of both terms so pos1 and -3 like that or pos1 with the x and -3 with the 2x i'm just going to check so here I've guessed I've written four guesses out and now I'm going to check that the inner and outer i'm going to multiply them and add them together and see if they add up to this middle term - 255x so here we have -1x and on the outside we have positive 26x clearly those add up to positive 25 not - 255 but it also gives us a clue as to which one will work since we got 25x and we want - 255x we just need to change the signs of those numbers so that means this one should be the one that works 1x and -26x adds up to - 255x so this is the combination that works if you tried these others they will not work you have -3x and pos2x which adds up to -1 and on this one you'd have 13 - 2 which would add up to 11x so the correct factorization is the binomial 2x + 1 * the binomial x -3 here is another factoring problem in which a does not equal 1 it equals 3 so we don't have any GCF here to pull out so we're stuck with either using guess and check slide and divide or grouping i'm going to go ahead and do this one by slide and divide so by slide and divide you take the three off of the first term and multiply it by the last term but don't worry too much about that we will end up dividing that out at the end we next ask ourselves what are two numbers that multiply to be 63 that will add up to 17 and your factors of 63 are 1 * 63 3 * 21 or 7 * 9 none of these if you were to add them add up to 17 so that's what tells you this polomial is a prime polomial which means it cannot be factored this next problem is asking for us to factor it by grouping grouping is always the method we want to use when we have four terms although we could have added these two together to create a tromial and factor it that way but they're specifically asking us to use the grouping method when we factor by grouping we're trying to pull the GCF out of these two terms and then the GCF out of these two terms as a first step the GCF of the first two terms is simply x and x * 4x would give us 4x^2 x * -3 would give us -3x whenever there's a minus here on these two terms we need to factor out a minus and the GCF of 12x and 9 is three so essentially we're factoring out -3 so we could divide both of these two terms by -3 to find what goes here which is 4x - 3 cuz -3 * 4x is -12x and -3 * -3 is pos9 now for something to be factorable by grouping these binomials must match which they do so our final answer is 4x - 3 we pull that out of both of these and these last two factors drop down into a new set of parentheses so this is the completely factored answer and we did the problem by grouping you could always check this so 4x * x is 4x^2ar 4x * -3 is -12x -3 * x is -3x and -3 * -3 is pos9 the order is just switched on these but that doesn't mean anything these answers are equivalent number 33 has three parts the first part is to find two numbers whose product is the 8 * 5 or 40 that adds up to 14 so you got to think of what are all the multiples of 40 so we have to start with the multiples of 40 what multiplies to be 40 1 * 40 2 * 20 4 * 10 5 * 8 but we have the pair already 4 + 10 would be 14 and four time 10 is 40 so letter A is just asking us what those two numbers are that would multiply to be 40 and add up to 14 and those two numbers are 4 and 10 part B is asking us to write 14x as two terms using the values 4 and 10 we would just change the 4 and 10 to 4x + 10 x for part C they're asking us to factor the problem by grouping so that is why they had us go through this process is because we want to rewrite the original problem changing 14 x into 4x + 10 x then we will factor out the GCF of the first two terms which is 4x leaving us with 2x + 1 and the GCF of the last two terms which would give us 2x + 1 so the final answer after pulling out the 2x + 1 is 2x + 1 * 4x + 5 so you might have been able to find out those factors using a different method but for this problem they're wanting us to go through all three steps here's another factoring problem you might recognize this as the difference of two perfect squares a square minus b^ 2 always factors into a + b a minus b if we take the square of 169x^2 and the of 9 which is 13x and 3 then those will be our a and b's that we plug into this formula so our final factored answer would be 13x + 3 * 13x - 3 feel free to multiply that out to check it but that should work the next question is another factoring problem it is of the form of a difference of two perfect squares so if we take the square root of both of these terms we would get 9 r^ 2 and 16 those are the a and b that we want to plug into the difference of two perfect squares formula so again a^ 2us b^ 2 factors into a + b a minus b we should notice that we have another difference of two perfect squares here so this one will go further into 3 r + 4 and 3 r -4 so the final answer is going to be 9 r 2 + 16 * 3 r + 4 * 3 r - 4 keep in mind there is no sum of two perfect squares formula so the 9 r^2 + 16 cannot be factored further so this is a final answer this next problem has already been factored and they just want us to solve for x recall that we can use the property that if a * b equals 0 then either a equals 0 or b equals z although this is an or statement this typically leads to more than one answer so in our problem 4x * x - 3 = 0 so either 4x = 0 or x - 3 = 0 dividing both sides by 4 we get our first solution of 0 adding three to both sides we will get our next solution of x= 3 we could check these by substituting them into the original problem if you substitute 0 right here you would get 0 * x - 3 and 0 * anything is 0 so that one checks then we could substitute 3 into this expression here so 4x * 3 - 3 is 0 so 4x * 0 is also zero so both of these check these are our two answers this problem has already been factored as well so we can set both of the factors equal to 0 to get our two solutions 7x - 2 = 0 or 3x + 10 = 0 adding 2 to both sides we get 7 x = 2 and dividing by seven we get x = 27th as one of our answers for the next problem if you subtract 10 from both sides we get 3x = -10 and dividing by 3 we get x = -10/3 so those are our two solutions this problem has already been factored on the left side however since the right side does not equal zero then that factorization is not going to be useful so we need to multiply this through x * 2x is 2x^2 and x * -7 is - 7x then we will subtract 39 from both sides subtracting 39 to both sides allows us to move that term over and set it equal to zero at this point we want to factor it but the original problem is asking us to solve so once we get it factored we will set our factors equal to zero to solve the problem we know that 2x * x is 2x^2 and for -39 we have a few choices i think 1 and 39 are going to be too far apart to add up to -7 even when the one's being multiplied by the two i'm going to focus on 3 and -3 or -3 and positive3 and because I want them to add up to the -7 I'm thinking I want the -3 here and the positive3 here let's go ahead and check it and see if that works 2x * x is 2x^2 and -3 * 3 is -39 so we just need to make sure the inner 13x and the outer pos6x adds up to -7x and it does so because that checks that's our correct factorization we're going to solve it by setting our factors equal to zero the second one if you subtract three from both sides we get -3 the first one if we add 13 to both sides we get 2x = 13 so dividing by two we will get 13 halves as one of our answers and -3 is the other answer next we're going to switch over to radicals this problem is asking us to simplify the square root of 192 so you can first type it into your calculator square 192 to see if it's a perfect square but it isn't so we need to instead break this down using our perfect square list like 2^2 is 4 3 squared 9 4^2 16 5 2 6 2 36 so that list obviously is infinite but I'm going to see if I can find a number here that goes into 192 evenly so I found that if I divide by 16 I will get 12 so but 12 has a perfect square also so I think that we have a larger perfect square that goes into this uh 64 that looks like it will get us directly to our answer if you divide 192 by 64 you get three so 64 * 3 and 64 is 8 so our final exact answer here is 8 * 3 this next question is asking us to simplify the fraction square of 125 over 25 the first thing I would check is to see if 25 divides into 125 which it does five times we can actually do the division first and get square five well five has no perfect square factor so square 5 is a final answer the next problem's asking us to simplify this expression recall that we want to make sure that every radical is broken down first and then we need to look for like radicals because those are like terms that we can add or subtract and combine into another term square 3 is already simplified and square 5 is already simplified so we can next jump to looking for our like terms and we have a 6 3 a 9 3 and a minus3 3 so we're going to want to combine all of those and then we have a square 5 that will just end up in our final answer when we are simplifying these we want to look at the numbers in front 6 + 9 is 15 - 3 is 12 12 3 + 5 and that's the final answer looking at problem 42 we see that these radicals aren't simplified yet and 9 goes into 45 five times so we could change square 45 to 9 * 5 for the second term we can change square 8 into square 4 * 2 we'll also want to break down 20 using 4 * 5 and 18 is 9 * 2 so next I'm going to perform those square roots so 7 * 3 * 5 - 3 * 2 * 2 - 2 * 5 + 3 * 2 Next I'll simplify the numbers in front of the square roots by multiplying them 7 * 3 is 21 5 we have a -3 * 2 so -6 2 - 2 5 and + 3 2 next we're going to look for our like terms we have a couple square 5 terms here 21 - 2 gives us 19 5 and then we have some square2 terms here -6 + 3 is - 3 2 so that's a final answer number 43 has two different radicals being multiplied together there's different ways to approach these but since 18 * 4 isn't too big of a number I'm just going to multiply them first all under one radical so 18 * 4 is 72 and p 7th * p 1st is p 8 so basically what I use there is that square a * b is the same as square of a next I want to work on breaking these down by pulling out any perfect square factors so 72 is 36 * 2 and the fastest way to determine when you have an even number here to determine what comes out is to divide the power by the root the index so 8 / 2 would give us p 4th that comes out 6 2* p4th and the p 4th would get multiplied in front leaving us with a final answer is 6 p 4th* 2 there are other ways to see how that happens you can change the square of p to the 8 square of p^ 2 * p^ 2 * p ^ 2 * p ^ 2 any of these square p^ 2 would be a p coming out so p * another p * another p * another p which is why we have p to the 4th out in front for this problem we have a radical multiplied by parentheses with two radicals inside so we can distribute that square 10 10 * 5 is 50 and 10 * 2 is 20 next we can work on breaking down those square roots by dividing out perfect square factors 25 is the largest perfect square that divides into 50 with no remainder square 4 divides into 20 5 * taking the square root of those perfect squares we get 5 2 + 2 5 which is the final answer to this question here we have two terms times two terms and both of them have a radical in them if you notice that this is of the form a + b * a minus b then you would know that the answer is a^ 2 - b^ 2 without having to do four multiplication problems but if not you could always foil this to get to the final answer i'll just do it the shortcut way which is squarex squared square and square root cancel each other so the first term of the answer is x the last term will be found by 7 * -7 or - 7^2ar which is -49 so after multiplying these out you should get x - 49 even if you used foil and I could show you how that works. x + 7 * x - 7 these two will equal x the outer is -7 x the inner is positive 7 x which will add up to 0 here and then 7 * -7 is - 49 so you still get x - 49 even if you multiply it out this problem is in the form of a binomial squared where the binomial part has a radical in it a minus b^ 2 = a^ 2 - 2 a b + b^ 2 or we could write this out as x - 6 * x - 6 and do the four multiplication problems i'll just go ahead and multiply it out so square x * x is x^2 which is simply x -6 * x is the next term and then another -6 * x and -6 * -6 is pos 36 these two add up to -12 x and so the final answer is this right here if you did the binomial squared method you would get x^2 which is x you would get -2 * the a which is x * the b which is 6 and that would be -2 x and then you would get + b^2 which is + 6^2 or 36 for this problem we want to look at simplifying 27 m 9th / 3 m to the 5th it might be helpful to separate these into 27 over 3* m 9th over of m 5th 27 is simply 9 * 3 so this 3 factor can be cancelled leaving us with 3 for the numerical part of this answer for the variables we could bring this into one fraction and then simplify by subtracting to get square of m to the 4th and m^2 * m^2 is m 4th therefore the final answer is 3 m^2 now a lot of these problems they say up front that the variables are positive numbers and that's how we can simplify it just thinking of the positive roots this problem is asking us to rationalize the denominator which simply means to get the radical out of the denominator and since it's not a perfect square we need to multiply by that same thing over itself in order to get that out of the denominator now realize that 2 w / 2 w is just a form of 1 so we're allowed to multiply by a form of one without messing up the value of this fraction when we multiply straight across we will just get square of 2 w we will get square of 2 w^ squared but the square and square root will cancel so we'll just get 2 w over 2 w you might be tempted to cancel the twos and the W's but that's square 2 versus two and square W versus W so they're not the same thing this would be the final answer this next problem is also asking for us to rationalize the denominator so we can use the property that square a over b is the same as a over b so to start with we'll just rewrite this as 11 over of y next we will want to rationalize the denominator by multiplying by that denominator over itself 11 * y is 11 y and y * y is y^ 2 which is simply y so that's the final answer those y's cannot be cancelled because in the numerator we have square y and in the denominator we have y then we move on to radical equations in order to solve an equation with a radical we first need to isolate the radical which right now it's the only thing on the left side so that step's been done for us then we want to reverse the radical by going to the power that index this is an invisible two so we are going to square both sides the square and square root will cancel and we'll get x + 14 = 9 subtracting 14 from both sides we get the final answer of x= -5 we should always check these answers square of x we think is -5 + 14 we're seeing if that would equal 3 so -5 + 14 is 9 and 9 is 3 so we know our answer is correct this next problem also involves solving an equation with a radical in it but this time the radical has not been isolated yet so we need to get x alone on this left side so we'll begin by adding 3 to both sides of the equation leaving us with x = 7 at this point we can get rid of that square root by squaring both sides and square and square root cancel each other out so we get x = 7 * 7 or 49 now again we should always check our answers so the check would look like 49 when we plug that in minus 3 and we're checking to see if that equals 4 the principal of 49 is 7 and 7 - 3 is 4 x= 49 is the correct answer number 52 is giving us a right triangle and asking us to solve for the hypotenuse since that's across from the 90° angle so basically they're asking us to use the Pythagorean theorem and that theorem is a^2 + b^2 = c^2 where a and b are the legs those are the two values that connect at the 90 and C is the hypotenuse which is the side across from the 90° angle so we can plug in 2 and 4 for A and B 2 + 4 2 and X is our C and we're trying to solve for X so before we can add these together we need to square both of them 2^2 is 4 4^2 is 16 then we add them together the last step here to get x from x^2 is to square both sides x^2 is x and square 20 is 4 * 5 which simplifies to 2 5 so that's our final answer for the length of the hypotenuse for number 53 it's another Pythagorean theorem problem where they're trying to have us solve for a side that is not the hypotenuse because across from the 90 is the hypotenuse and that's length 11 so this is our hypotenuse and these two other values are the legs so in terms of a^2 + b^2 = c^ 2 x is one of the a or b i'll just call it a 8 is the other one and C the hypotenuse is 11 so the first thing we're going to do is to square the 8 and square the 11 and then we need to isolate x2 by subtracting 64 from both sides so we will get x^2 = 57 so since x^2 = 57 the square root of x^2 is the square root of 57 and that has no perfect square factors so that would be our final answer so x= 57 is the length of the unknown side this next problem involves the square root property so what the square root property says is that if we have something like x2 = a then if we square root both sides we will get plus or minus so our answers would be plus or minus the square root of the other side we're going to get two answers to these problems so the first step is to square root both sides which will give us plus or minus on the side without the variable so w^2 the square root of that is w plus or minus we could treat this as 1 over 16 to make it easier to do in our head and 1 is 1 16 is 4 so this is two answers positive 1/4 and 1/4 for number 55 we will also need to use the square root property but first we want to isolate the part that's being squared which is x so we will divide by three first once we have these squared isolated we can square root both sides which leads to plus or minus answers they usually want us to rationalize the denominators so we need to multiply by 3 over 3 3 * 14 is 42 and 3 * 3 is 3 so you might want to ask yourself is there a perfect square factor of 42 think of your perfect squares 4 9 16 25 36 those are all the perfect squares that are less than 42 that we could divide by and none of these divide into 42 evenly so that's a final answer positive 42 over3 and negative 42 over3 number 56 has the other type of equation in which using the square root property is a quick way to get to the answer so we have this binomial squared here so we can square root both sides automatically since the binomial is isolated that will leave us with a plus or minus in front of the square two the square and square root effectively cancel each other giving us x - 3 = plus or minus 2 then when we add three to both sides of the equation we will get 3 + 2 and 3 - 2 as our two solutions this next problem involves simplifying in order to solve a quadratic equation so remember your quadratic formula is that x =b + or minus b^2 - 4 a c all over 2 a and they're talking about the form of a quadratic equation ax^2 + b x + c = 0 so in this case a is 4 b is 7 c is -1 so when you plug those in opposite of b which is -7 plus or minus square of b^ 2 - 4 * a which is 4 * c which is -1 all over 2 * a which is 2 * 4 so we just need to simplify this problem -7 + or - of 7 * 7 is 49 and -4 * 4 is -16 * -1 is pos6 and that's all over 8 so simplifying further we can add the 49 + 16 to get 65 and that's all over 8 so the final simplified answer is -7 + or minus 65 all / 8 the next problem is asking us to graph a quadratic equation it's quadratic because we have x squared here remember when the highest degree of the polomial is a two then it's a quadratic equation and those graph into a parabola so typically your parabola is like this or like this okay so if we can find the vertex that will be very helpful and then finding the axis of symmetry helps as well because that will divide the left and the right sides of the parabola so this particular problem is asking us to solve for some ordered pairs so it's giving us -2 -1 0 1 and 2 so they want us to plug in all those for x and simplify all this in order to get the corresponding y value so in other words for the first problem y = -4 * -2^ 2 -2 is 4 * -4 would be -16 the second problem y = -4 * -1^ 2ar so -4 * 1 is -4 when x is 0 y would also be zero because so 0 * -4 when x is 1 we can multiply 1 * -4 so we have two -4 values and that's a good thing when you have two y values at the same height because then you can figure out that your axis of symmetry will be right between these two xcoordinates and halfway between those two x coordinates the midpoint of -1 and 1 would be zero so we know this will actually be the vertex and that's important in graphing the parabola so whenever you see two y values that are the same height essentially you can look at the x values and find the number right in between them and that will be the xcoordinate of the vertex which means this is the y-coordinate of the vertex now let's solve for this last value -4 * 2 ^ 2 -4 * 4 is -16 so I'm just going to graph a couple of these um 0 0 again is going to be our vertex and we could see that anyway when we plot a few of these like 1 -4 is down here and -14 is here so I'm not going to plot the -2 -16 and 2 -16 because those will be going off my graph but let's go ahead and connect our parabola basically in this case your axis of symmetry is the y ais for this quadratic equation they're asking us to solve for all the intercepts and graph if we plug in 0 for x we can find the y intercept so our y intercept is going to be found by y = 0^ 2 - 6 * 0 + 10 which is 10 the coordinates of our y intercept are 0 comma 10 remember you solve for x intercepts by setting y= 0 and solving for x and you solve for y intercepts by setting x equal to zero and solve for y if we wanted to set y equal to zero now to solve for x we would get this equation but that's not factorable since -2 * -5 aren't going to add up to -6 and neither are -1 and -10 so it's not factorable in order to solve this we would probably want to use the quadratic formula so x = b + or minus b^ 2 - 4 a c all over 2 a the opposite of b is opposite of -6 plus or -6^ 2ar - 4 * 1 * 10 so you'll have 36 - 40 under the square root which will give us imaginary answers so because these are going to be imaginary they're not going to help us with our graph so at this level we're just going to say these are imaginary and I'm not going to go ahead and I'm not going to solve those any further so basically what that tells us when our when our solutions are imaginary is that our parabola will not intersect the xaxis so it's either this type where you have a parabola above the x axis facing up or it's this type where you have a parabola below the xaxis facing down because it's not intersecting the xaxis a key point is the vertex so let's solve for the vertex now the vertex can always be found by an xcoordinate ofb / 2 a and then plugging that in for x which is called f of it so we'll get this value then we'll plug that value in for x and solve for y and that'll give us the coordinates of the vertex what is b well if b is -6 then b is -6 which is 6 the xcoordinate of the vertex is b / 2 a which is -6 over 2 * 1 so we're using these two values a and b we're plugging them into the vertex formula and we get 6 / 2 which is 3 that tells us the xcoordinate of the vertex is three so how do we get the ycoordinate well we plug that in that's what f of that number means you plug in the three for x so the ycoordinate of the vertex is going to be at 3^ 2 - 6 * 3 + 10 and that would be 9 - 18 which is -9 + 10 which is 1 so our coordinates of our vertex are 3 1 so next let's go and graph this parabola that will be a point on the parabola 010 that's our y intercept and 31 is the vertex 1 2 3 1 is the vertex so that's another point on the parabola and keep in mind the axis of symmetry always goes through so this is not part of the graph it's just what we reflect over to get another point on the graph so we can reflect this point that's up here at 10 all the way across to get a new point and if you say well what how do we know what that new point is well from our vertex we could get to this point 10 by going left three and up 10 so if we go right three and up 10 we will get to another point on this parabola so let's go ahead and graph the parabola now I'll do that in purple and of course these should be smooth U-shaped curves but that's just a rough sketch for this next quadratic equation they want us to solve for the vertex and the intercepts remember we can solve for the y intercept by setting x= 0 so y = 2 + 10 * 0 - 9 which is -9 so the y intercept is found at 0,9 to find the x intercepts we can set y equal to zero and this problem looks like it might be factorable so I'm going to factor out -1 in order to make my x term a positive inside the parentheses and then I notice that it is actually factorable x - 9 * x -19 * -1 is pos 9 and -9 + -1 is -10 setting all three of these factors equal to 0 we can see that this is an extraneous solution 0 = -1 is false you can add 9 to both sides and add one to both sides so this is giving us some intercepts when x is 9 y was 0 and when x was 1 y was 0 and the third intercept is 09 okay so you have your two x intercepts here and again those are found by setting y equal to 0 and solving for x you could still plug those into the quadratic formula i just found it a little bit easier to factor this one since it's factorable next they wanted us to find the vertex so the vertex is always found the xcoordinate of it is found by over 2 a that's the formula since a is 1 b is 10 so -10 / 2 * -1 -10 / -2 is 5 so the x coordinate of the vertex is five we just need to find the ycoordinate of the vertex by substituting 5 into the original problem so y = -x^2 means 1 * x^2 our x that we're plugging in here is 5 + 10 * 5 - 9 so we get - 255 + 50 - 9 and 25 - 9 is 16 since it's a negative that means it opens down since a is a negative number it's -1 our vertex is over five up 16 these are our two x intercepts the ycoordinates are the same height then right in between the x coordinates so halfway between 9 and one meaning five is where the vertex is found plugging in that five for x you get the y-coordinate of the vertex so we have four points on that parabola and that would be pretty easy to graph so I just want to wish you good luck on your final exam if you found this video helpful please like and subscribe feel free to comment i appreciate it