# What is the polynomial equation for the points passing through (-2,0), (0,0), (1,-4), (2,0)?

Apr 6, 2016

$f \left(x\right) = \frac{4}{3} \left({x}^{3} - 4 x\right)$

#### Explanation:

First let defined a polynomial function in factor form as follow

$f \left(x\right) = a \left(x - b\right) \left(x - c\right) \left(x - d\right) \ldots . . \left(x - z\right)$,

where $a$ is a none zero leading coefficient and

$b , c , d \ldots . . z$ are the$x$ intercepts, of the function, which meant
$x = b , x = c , x = d \ldots \ldots x = z$

We are given the following x- intercepts

$x = - 2 , y = 0$
$x = , y = 0$
$x = 2 , y = 0$

And the value of
$x = 1 , y = - 4$

We can re-write the x-intercepts in the factor forms like this

$f \left(x\right) = \left(x\right) \left(x + 2\right) \left(x - 2\right)$

the general form is

$f \left(x\right) = a \left(x\right) \left(x - 2\right) \left(x + 2\right)$

f(x) = a(x(x^2-4)

f(x) = a(x&3 - 4x)

Since $x = 1 , y = - 4$ , we can substitute that into the general form to solve for $a$

$f \left(1\right) = a \left({1}^{3} - 4 \left(1\right)\right)$
$- 4 = a \left(1 - 4\right)$
$- 4 = - 3 a$
$a = \frac{4}{3}$

So the general polynomial function that passing through the points
$\left(0 , 0\right) , \left(2 , 0\right) , \left(- 2 , 0\right) , \left(1 , - 4\right)$ is

$f \left(x\right) = \frac{4}{3} \left({x}^{3} - 4 x\right)$

$f \left(x\right) = \frac{4}{3} {x}^{3} - \frac{16 x}{3}$