# How do you write a quadratic function in intercept form whose graph has x intercepts -7, -2 and passes through (-5, -6)?

Oct 9, 2017

$y = \left(x + 7\right) \left(x + 2\right)$

#### Explanation:

$\text{given the x-intercepts ( zeros) of a quadratic function}$

$\text{say "x=a" and } x = b$

$\text{then the factors are "(x-a)" and } \left(x - b\right)$

$\text{and the quadratic function can be expressed as a}$
$\text{product of the factors}$

$y = k \left(x - a\right) \left(x - b\right)$

$\text{where k is a multiplier and can be found if we are}$
$\text{given a point on the parabola}$

$\text{here "x=-7" and } x = - 2$

$\Rightarrow \left(x + 7\right) , \left(x + 2\right) \text{ are the factors}$

$\Rightarrow y = k \left(x + 7\right) \left(x + 2\right)$

$\text{to find k substitute "(-5,-6)" into the equation}$

$- 6 = k \left(2\right) \left(- 3\right) = - 6 k \Rightarrow k = 1$

$\Rightarrow y = \left(x + 7\right) \left(x + 2\right) \leftarrow \textcolor{red}{\text{ in intercept form}}$