Find the quadratic function y = ax^2 + bx + c whose graph passes ...

Question

Find the quadratic function y = ax^2 + bx + c whose graph passes through the points (1, 4), (3, 20), and (-2, 25).

Accepted Answer

Hello today we're going to be solving for the coefficients of a given equation. So the equation given to us is y. Is equal to P. X squared plus Q. X plus R. And we are told that it passes through the points negative +3003 and +10. So how do we start solving for the coefficients of this equation? While what we can do is go ahead and take our points and directly plug them in to our equation. That way we can try to see if we can make a system of equations. So let's go ahead and start by plugging in our force. Our first point which is negative three comma zero. What we're gonna do is we're taking this X. Value and replacing it for every X. And we're taking this Y value and replacing it for Y. Doing so is going to give me negative three squared P plus negative three times Q plus R. Is equal to zero. And just to simplify this negative three squared is going to be positive nine. So we have nine P positive negative three is gonna give me negative three Q Plus R. And that is going to equal to zero. And we're gonna allow this to be our first equation. Now what we're gonna do next is plug in our next point. So our next point is zero comma three. And just like before we're replacing every X. And everyone in our equation with the given point doing so is going to give me zero squared P plus zero times Q plus R. Is equal to three. Well zero squared is just +00 times P is going to cancel itself out and zero times Q is gonna cancel itself out. So what we are left with is R is equal to three. And we can go ahead and label this as our second equation. But this is also one of our solutions because we have a value for our now. Now finally let's plug in our last point which is 1:00. Plugging Our last point is going to give me one squared times P plus one times Q plus R is equal to +01 squared is just one and one times P is just P one times Q is Q plus R is equal to zero. And we're going to allow this to be our third equation. Okay, so we don't necessarily have a system of three equations. We have a system of two equations because our first point and our third point gave us a system of equations of three variables. But our second point gives us a value for our so what we can do is go ahead and plug in our value for both of our equations to try to simplify it. So plugging in our for our first equation, What we get is nine p minus three Q Plus three because our S three equal to zero. And again we want to solve for P and Q. So in order to do that, we're gonna subtract three to both sides of this equation. Doing so is going to give me nine, P minus three Q. Is equal to negative three. And I'm just gonna go ahead and label this as equation for now let's go ahead and do the same thing for equation three plugging in our two, equation three is going to give us P plus Q plus three equal to zero and just like before we want to sell for P. And Q. So we're gonna subtract three from both sides of this equation, leaving us with P plus Q equal to negative three And I'll go ahead and label this as equation five. Now what we wanna do is go ahead and use equations four and five and set it up as a system of equations to solve for P and Q. So continuing that here, equation four is nine p minus three Q equal to negative three. And our fifth equation was P plus Q equal to negative three. So what we wanna do is go ahead and eliminate one of the variables in order to do that. What we could do is go ahead and multiply our second equation here by three and doing so is going to give me nine p minus three Q. Equal to negative three. And our second equation is going to be three, P plus three Q. Is equal to -9. Now let's go ahead and add these equations together and simplify negative three Q plus three Q. Is going to cancel each other out. Nine. P plus three P is going to give us 12 P. And negative three minus nine is going to give me negative 12. Now in order to sell for P I'm gonna divide both sides by 12 and P is going to equal to negative one which is one of our solutions. So now we have a substitution for P. And R. What we can do is take these two values and plug it into our third equation to solve for Q. So plugging in P and are into our third equation is going to give us negative one plus Q. Plus three is equal to zero. Now combining like terms negative one plus three is gonna give me positive two. So what we have is Q plus two equal to zero. And subtracting two from both sides of the equation is going to give me Q equal to negative two. And that's gonna be our final solution. So just putting our solutions together, what we have, What we have is P is equal to negative one. Q. is equal to -2 and R. is equal to three. And that's gonna be the set of solutions for P. Q. And R. And that also means that the answer to this problem is going to be D. So I hope this video helps you in understanding how to find the coefficients of an equation and I'll go ahead and see you all in the next video.

Find the quadratic function y = ax^2 + bx + c whose graph passes through the points (1, 4), (3, 20), and (-2, 25).

Key Concepts

Quadratic Functions

Systems of Equations

Substitution Method

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