# How do you write a quadratic equation with x-intercepts: -4,-2 ; point: (-6,8) ?

Oct 9, 2017

$y = {x}^{2} + 6 x + 8$

#### Explanation:

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

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

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

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

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

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

$\text{here "x =-4" and } x = - 2$

$\Rightarrow \left(x + 4\right) \text{ and "(x+2)" are the factors}$

$\Rightarrow y = k \left(x + 4\right) \left(x + 2\right)$

$\text{to find k substitute "(-6,8)" into the equation}$

$8 = k \left(- 2\right) \left(- 4\right) = 8 k \Rightarrow k = 1$

$\Rightarrow y = \left(x + 4\right) \left(x + 2\right) \leftarrow \textcolor{red}{\text{ in intercept form}}$

$\text{distributing and simplifying gives}$

$y = {x}^{2} + 6 x + 8 \leftarrow \textcolor{red}{\text{ in standard form}}$