# How do you write a quadratic function in intercept form who has x intercepts 4, 6 and passes through points (5,-2)?

Aug 25, 2017

$y = 2 \left(x - 4\right) \left(x - 6\right)$

#### Explanation:

$\text{given x-intercepts "x=b" and } x = c$

$\text{we can express the factors as "(x-b)" and } \left(x - c\right)$

$\text{and the quadratic function as the product of the factors}$

$\Rightarrow y = a \left(x - c\right) \left(x - b\right) \leftarrow \textcolor{b l u e}{\text{ a is a constant}}$

$\text{here the intercepts are "x=4" and } x = 6$

$\Rightarrow y = a \left(x - 4\right) \left(x - 6\right)$

$\text{to find a substitute "(5,-2)" into the equation}$

$- 2 = a \left(1\right) \left(- 1\right) = - a \Rightarrow a = 2$

$\text{the quadratic function is therefore}$

$y = 2 \left(x - 4\right) \left(x - 6\right) \leftarrow \textcolor{red}{\text{ in intercept form}}$
graph{2(x-4)(x-6) [-10, 10, -5, 5]}