# How do you write a quadratic function in intercept form whose graph has x intercepts -5, 5 and passes through (6, 11)?

Apr 6, 2017

Quadratic function is $y = f \left(x\right) = {x}^{2} - 25$

#### Explanation:

As the quadratic function, say $y = f \left(x\right)$ has $x$-intercepts $- 5 , 5$

it's value is $0$, when $x = 5$ and $x = - 5$.

Hence equation must be of the type

$y = f \left(x\right) = a \left(x - 5\right) \left(x - \left(- 5\right)\right) = a \left(x - 5\right) \left(x + 5\right)$

as it passes trough $\left(6 , 11\right)$, we have

$11 = a \left(6 - 5\right) \left(6 + 5\right) = a \times 1 \times 11 = 11 a$

Hence, $a = 1$ and quadratic function is $y = f \left(x\right) = \left(x - 5\right) \left(x + 5\right) = {x}^{2} - 25$