Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer

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

So you have to solve following system of equations :

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.