Getting a standard form quadratic from a set of points ($3$ points)

Question

I came home from school today, pulled out my homework, now I'm stumped. I don't want the answer, I just want to know how to do it.

Here is the question that I'm reading:

Determine a quadratic function in standard form for the set of points $(x_1,y_2)(x_2,y_2)(x_3,y_3)$

Where the $x$'s and $y$'s represent the points.

Here are the points I am working with (remember, I don't want the answer, I just want to know how to do it, there are lots of questions on this so I want to be able to know how to do it).

$(-1, 2), (0, 1), (1, -4)$

Thanks!

Answer 1

Instead of solving a system of equations, you can Lagrange interpolation, since we know that the sum of quadratic polynomials is itself a quadratic polynomial. The strategy is to express the solution as a sum of three polynomials:

• $p_1(x)$ which passes through $(-1,2)$ and has roots at $0$ and $1$.
• $p_2(x)$ which passes through $(0,1)$ and has roots at $-1$ and $1$.
• $p_3(x)$ which passes through $(1,-4)$ and has roots at $-1$ and $0$.

In each line, the roots are at the $x$ coordinates of the two points we're not hitting. So when we add all three partial solutions, $p_1+p_2+p_3$ will pass through all three points.

For $p_1$, we start by constructing a quadratic with roots at $0$ and $1$ and then scale it such that it has the right value at $-1$. To get roots at $0$ and $1$ we just multiply $x-0$ and $x-1$ to get $x^2-x$. The value of $x^2-x$ at $x=-1$ is $2$, so we don't even need to scale it: $p_1(x)= x^2-x$.

For $p_2$, start by multiplying $x-(-1)$ and $x-1$ to get $x^2-1$. Its value at $0$ is $-1$, so we need to scale by $-1$ to get $p_2(0)=1$. Therefore $p_2(x)=-x^2+1$.

For $p_3$, multiply $x-(-1)$ and $x-0$ to get $x^2+x$. The value at $1$ is $2$, so we scale by $\frac{-4}{2}=-2$ and get $p_3(x)=-2x^2-2x$.

Now add them all together.

Answer 2

Standard form is :$f(x)=y=ax^2+bx+c$

So you have to solve following system of equations :

$\begin{cases} y_1=ax_1^{2}+bx_1+c \\ y_2=ax_2^{2}+bx_2+c\\ y_3=ax_3^{2}+bx_3+c \end{cases}$