# How do I find the x-intercepts of a quadratic function in vertex form (x+4.5)^2-6.25?

May 24, 2015

For a quadratic equation of the form
$y =$ and expression in $x$

The x-intercepts occur at the points where $y = 0$

Given ($y =$)$\left({\left(x + 4.5\right)}^{2} - 6.25\right)$
We need to solve
${\left(x + 4.5\right)}^{2} - 6.25 = 0$

${\left(x + 4.5\right)}^{2} = 6.25$

$x + 4.5 = \pm 2.5$

$x = - 4.5 \pm 2.5$

The x-intercepts are
$\left(- 2\right)$ and $\left(- 7\right)$