In Exercises 19–22, find the quadratic function y = ax^2+bx+c who...

Question

In Exercises 19–22, find the quadratic function y = ax^2+bx+c whose graph passes through the given points. (−1, 6), (1, 4), (2, 9)

Accepted Answer

Hello. Today we are asked to find the quadratic function that passes through the given points. So in order to figure this out, what we can go ahead and do is plug in the given points into our unknown equation. And we can go ahead and use those points to set up a system of equations of three variables. So let's go ahead and start by plugging in two comma three into the unknown quadratic two is in the exposition and three is in the white position. So plugging this in is going to give us a times two squared plus B. Times two plus C is equal to y, which is three. Now two squared is going to give me four. So that's going to give me four A two times B is going to give me to be so I have to B plus C is equal to three. And I'm gonna go ahead and let this be equation one and we're going to repeat this process for the other given points. Now we're gonna go ahead and plug in negative one comma one into the unknown quadratic and plugging this in is going to give us a times negative one squared plus B. Times negative one plus C is equal to Y. Which is one here negative one squared is going to give me positive one and one times A is just going to give me a negative one. Times B is going to give me negative B plus C is going to equal to one. And this is going to be equation too. Now we're gonna go ahead and plug in our last point, which is 1:02 plugging this in is going to give me a times one squared plus B. Times one plus C is equal to Y, which is +21 squared is going to give me one and eight times one is just going to give me a B times one is going to give me B plus C is equal to and this is going to be our third equation. Now we're going to use the three equations we just came up with and set them up as a system of equations in order to sulfur A. B and C. Now what I want to go ahead and do is pick any combination of the three equations and cancel out one of the variables to make another equation of two variables. That might sound a little confusing for now, but let's just go ahead and jump right into it. I want to go ahead and add equation two and 3 together. So equation two is going to be a minus B plus C is equal to one and equation three is going to be a plus B plus C is equal to two. If you notice right away, adding these two equations together is going to cancel out the B term and that's going to leave us with A plus A which is two, A plus C, times C plus C, which is two C is equal to two plus one, which is three. What I've essentially done is come up with another equation of two variables involving A. And C. So I'm gonna go ahead and call this equation for for now now we're going to repeat this process but with a different combination of equations and I'm gonna go ahead and and add equation one with the equation too, equation one is four A plus two. B plus C is equal to three. An equation two is a minus B plus C is equal to one. And I want to go ahead and add these together in a way that creates another equation of two variables. But since the fourth equation involves A. N. C. I want to go ahead and cancel out be from equation one and two. And in order to do that I can go ahead and multiply equation two x 2. And doing this is going to give me four A plus two. B plus C. Is equal to three and we're going to be adding to a minus two. B plus two C. Is equal to two. If we add these equations together the B. Is going to cancel each other out and what we are left with is four A plus two A. Which gives me six A. Plus two C plus C. Which gives me three C. Is equal to three plus two which is five. And this is going to be equation five. Now what I want to go ahead and do is add equation four and five together and cancel out one of the variables either A. N. C. So that way we can solve for at least one of the variables adding equation four and five together is going to give me two A plus two C. Is equal to three and add an equation five is going to give me six A plus three C. Is equal to five. I'm gonna go ahead and cancel out the A term. And in order to cancel out the A term I'm going to multiply equation four by negative three. Multiplying equation for by negative three is going to give me negative six A minus six C. Is equal to negative nine. And we're going to be adding six A plus three C. Is equal to five. Once we add these equations together the A variable is going to cancel itself out and that leaves me with negative six C plus three C. Which is going to give me negative three C. Is equal to negative nine plus five which is going to give me negative four. And in order to sulfur. See I'm going to divide both sides of this equation by -3. And that means that C. is going to equal to positive 4/3 because dividing the negative number by negative number is going to produce a positive number. So we have a value for C. Now what I'm going to do is plug this into either equation four or five and we can go ahead and sell for a. I'm gonna go ahead and plug this in into equation five, plug it in our C. Value to equation five is going to give me six A plus three times C. Which is 4/3 is equal to 53 times 4/3 is just going to cancel out the denominator and what we are left with is six A plus four is equal to five. Now I want to go ahead and sell for A and in order to do that I'm going to subtract four to both sides of the equation and that's going to give me six A. Is equal to five minus four which is positive one. And in order to solve for a I'm going to divide both sides of this equation by six and A. Is going to equal to 1/6. Now that we have a value for A and C. We can go ahead and plug this into one of the original equations. That way we can go ahead and solve for B. And I'm gonna go ahead and plug this into equation three, plugging this into equation three is going to give me A which is 1/ plus B plus C, which is 4/3 Is equal to two. Now I want to go ahead and combine the fractions together, 1/6 and 4/3. So we're going to be adding 1/6 with 4/3. But in order to add these fractions together, we need to have a common denominator while the lowest common denominator between 1/ and 4/3 is six. So I'm going to multiply 4/3 by 2/2 and that's going to give me 1/6 plus 8/6 And this is going to give me 9/6 which is equal to 2/3. I'm sorry, 3/2. So what I have is B plus 3/ is equal to two. And in order to solve for B, I'm going to subtract 3/2 to both sides of the equation and subtracting 3/2 from two. Well, what we have is two minus 3/2 but we can put 2/1 and repeat this process of finding the lowest common denominator and the lowest common denominator between one and two is going to be two. So multiplying 2/1 by two is going to give me 4/2 minus 3/2, which is equal to 1/2. So B is going to equal to 1/2. Now that we have the values for A B and C. We can just go ahead and plug this in to our general quadratic and recall that the general quadratic was Y is equal to a x squared plus bx plus C and substituting in the values for a. B And C is going to give me Y is equal to 1/6 X squared plus B, which is 1/2 X plus C, which is 4/3. And this is going to be the quadratic equation that passes through the three given points of the problem. And that means that the answer to this problem is going to be C. So I hope this video helps you in understanding how to approach this type of problem, and I'll go ahead and see you all in the next video.

In Exercises 19–22, find the quadratic function y = ax^2+bx+c whose graph passes through the given points. (−1, 6), (1, 4), (2, 9)

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