# How do you write a quadratic equation with x-intercepts: -4,3 ; point: (-5,16)?

Sep 11, 2016

$y = 2 {x}^{2} + 2 x - 24$

#### Explanation:

Use the form $y = k \left(x - a\right) \left(x - b\right)$
where k is a constant, and a and b are the zeros (x coordinate of the x-intercepts).

$a = - 4$ and $b = 3$

$y = k \left(x + 4\right) \left(x - 3\right)$

$y = k \left({x}^{2} + 4 x - 3 x - 12\right)$

$y = k \left({x}^{2} + x - 12\right)$

To find k, plug in the given point $\left(- 5 , 16\right)$.

$16 = k \left({\left(- 5\right)}^{2} + \left(- 5\right) - 12\right)$
$16 = k \left(25 - 5 - 12\right)$
$16 = k \left(8\right)$
$k = 2$

Plug k into the equation found above.
$y = 2 \left({x}^{2} + x - 12\right)$

$y = 2 {x}^{2} + 2 x + 24$