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,−4), (1,−2), (2, 5)

Accepted Answer

Hello today we're going to be finding the quadratic function that passes through the given points. So the quadratic function given to us is why is equal to a X squared plus bx plus C. And we are told that the given points R negative one comma +71 comma 11. And native to comma 29. Now, in order to solve for the quadratic function, we need to go ahead and find the coefficients A, B and C. But how do we do that while we can go ahead and take our given points and plug it in to the quadratic function given to us and we can use that to solve a system of equations that will give us a B and C. So let's go ahead and take the first point which is negative one comma seven. And plug it into our giving quadratic since X is equal to negative one. This is going to give us a times negative one squared plus B. Times negative one plus C. And that's going to equal to y, which is seven negative one squared is going to give us positive one. And eight times one is just going to give us a B times negative one is going to give us negative B. And that we're gonna add C. And all of that is going to equal to seven. So this is going to be the first equation that we're going to use in our system of equations. Next let's go ahead and plug in one comma 11. Doing this is going to give us a times one squared plus B times one plus C is equal to 11. 1 squared is just going to give us one and eight times one is just a B times one is B. And we have C by itself and all that is going to equal to 11. This is going to be the second equation for a system of equations. And finally let's go ahead and plug in negative two comma 29. Plugging this into the quadratic is going to give us a times negative two squared plus B, times negative two plus C is going to equal to Y, which in this case is 29 negative two squared is going to give us positive four. So we have four A plus B times negative two which is negative two, B plus C is equal to 29. And this is going to be the third equation for the coefficients. So we have three equations involving three variables. So we can go ahead and solve this as a system of equations involving three variables. What we want to go ahead and do is pick a combination of any two of the three equations and add them together to cancel out one of the variables. That way we can go ahead and create another equation involving two variables instead. So let's just go ahead and add equation one and equation two together again, Equation one is a minus B plus C is equal to seven and equation two is a plus B plus C is equal to 11. Notice that if we add these equations together, the bees are going to cancel out because negative B plus B is going to give us zero, so A plus A is going to give us to a C plus C is going to give us to see and 11 plus seven is going to give us 18. Now notice that this equation each term has a coefficient of a factor of two, so we can divide this entire equation by two and this is going to give us A plus C is equal to nine. So what we've done is go ahead and create an equation involving two variables and I'm gonna go ahead and call this equation four. Next we want to go ahead and repeat this process but with a different combination of equations, let's go ahead and add equation one which is a minus B plus C equal to seven and add equation three to this. And this is going to be for a minus two, B plus C is equal to 29. Now, in order since we canceled out be in the first combination of equations, we want to go ahead and cancel out be in this combination of equations and we can do that by multiplying the first equation by -2. multiplying the first equation by negative two is going to give me negative two, A plus two, B minus two, C is equal to negative 14 and we're going to be adding equation three which is four A minus two B plus C. Is equal to 29. Now notice that to be minus two B. Is going to cancel itself out. So now we just need to simplify everything else for a -2. A is going to give me to a and negative two C plus C is going to give me minus C and that is going to equal to 29 minus 14 which is positive 15. Now I have another equation of two variables involving A. N. C. And this is going to be equation five. Now what we can go ahead and do is add equation four and five together that way we can go ahead and sell for A. N. C. So if we add equation four Which is a plus C is equal to nine with equation five which is to a minus C. Is equal to 15. Notice that if we add these equations together, the seas are going to cancel out. So that's just gonna leave us with a A plus to A. Is going to give me three. A and 15 plus nine is going to give me 24 in order to solve for a I'm just gonna divide both sides of this equation by three and A. Is going to equal to eight and that's the first very coefficient that we want to solve for. Now that we have a value for a we can go ahead and plug it into either equation four or five and we can go ahead and use that to solve for C. Let's go ahead and take this value of A and plug it into equation four. That's going to give us a plus C, or in this case eight plus C Is equal to nine. And in order to solve for C, I'm just gonna subtract 8 to both sides of the equation and that's going to give me C is equal to one, which is the next coefficient that we want to solve for now that we have the values of A. N. C. We can plug this into either equation one equation to our equation three. In order to solve for B, let's go ahead and plug this in to equation one, equation one is a minus B plus C, or in this case it's going to be eight minus B plus one and that's going to equal to seven. Eight plus one is going to give me nine, so I have nine minus B is equal to seven. And in order to solve for B, I'm going to subtract nine to both sides of this equation. Doing this is going to give me negative B is equal to negative two. And in order to solve for B I just need to divide both sides by negative one and B is going to equal to two. Now we have the coefficients for a B and C. What we want to go ahead and do now is plug this into the given quadratic. That was in the beginning of the problem. The given quadratic was Y is equal to a X squared plus b, X plus C. And plugging in our values of A B and C is going to give us Y is equal to eight X squared plus two X plus one. And this is going to be the the equation of the quadratic formula that passes through the three given points. And that means that the answer to this problem is going to be a. 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.

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In Exercises 19–22, find the quadratic function y = ax^2+bx+c whose graph passes through the given points. (−1,−4), (1,−2), (2, 5)