# 101fobster's question at Yahoo! Answers regarding finding a parabolic function

Staff member

#### MarkFL

Staff member
Hello 101fobster,

We are given 3 points through which the quadratic function must pass. Let's represent the function with:

$f(x)=ax^2+bx+c$

The 3 points we are given are:

(-1,0), (0,-10), (2,-6)

and so we may write the linear system:

$a(-1)^2+b(-1)+c=0$

$a(0)^2+b(0)+c=-10$

$a(2)^2+b(2)+c=-6$

or:

$a-b+c=0$

$c=-10$

$4a+2b+c=-6$

Since the second equation gives us $c=-10$, we now have:

$a-b=10$

$2a+b=2$

Adding these, we eliminate $b$ to obtain:

$3a=12\,\therefore\,a=4$

and so from the first equation:

$b=-6$

and thus:

$f(x)=4x^2-6x-10$

#### soroban

##### Well-known member
What is the quadratic equation with these points?
. . x-intercept: -1 . . y-intercept: -10 . . Point: (2,-6)

The parabola could "horizontal" . . . of the form: .$x \:=\:ay^2 + by + c$

The equation is: .$x \,=\,\text{-}\frac{1}{10}y^2 - \frac{11}{10}y - 1$

The problem did not specify a parabola.

The quadratic could be a circle: .$(x+\frac{17}{4})^2 + (y + \frac{43}{8})^2 \:=\: (\frac{5}{8}\sqrt{101})^2$

It has two x-intercepts: $\begin{Bmatrix}(\text{-}1,0) \\ (\text{-}\frac{15}{2},0)\end{Bmatrix}$ .and two y-intercepts: $\begin{Bmatrix}(0,\text{-}10) \\ (0,\text{-}\frac{3}{4}) \end{Bmatrix}$

I was surprised to see that all the coordinates are rational.