# How do you write a quadratic equation with x-intercepts: 0,2 ; point: (3,-6) ?

May 1, 2017

$y = - 2 {x}^{2} + 4 x$

#### Explanation:

The quadratic must have factors $x$ and $\left(x - 2\right)$ in order to have the required $x$-intercepts.

So it can be written in the form:

$y = k x \left(x - 2\right)$

for some constant $k$ to be determined.

Substitute the given point $\left(3 , - 6\right)$ to find:

$\textcolor{b l u e}{- 6} = k \left(\textcolor{b l u e}{3}\right) \left(\textcolor{b l u e}{3} - 2\right) = 3 k$

Divide both ends by $3$ to find:

$k = - 2$

So:

$y = - 2 x \left(x - 2\right) = - 2 {x}^{2} + 4 x$

graph{(4x^2+y^2-0.01)(4(x-2)^2+y^2-0.01)(4(x-3)^2+(y+6)^2-0.01)(y+2x^2-4x) = 0 [-3.6, 6.4, -7.08, 2.92]}