How do you find the x-intercepts of a quadratic function in standard form given f(x) = 2(x+1)^2 - 3?

Mar 29, 2018

$x = - 1 \pm \frac{\sqrt{6}}{2}$

Explanation:

This function is written in vertex form.
To find the x-intercepts, make f(x) = 0.
$2 {\left(x + 1\right)}^{2} - 3 = 0$
${\left(x + 1\right)}^{2} = \frac{3}{2}$
$x + 1 = \pm \frac{\sqrt{3}}{\sqrt{2}}$
$x = - 1 \pm \frac{\sqrt{3}}{\sqrt{2}}$, or
$x = - 1 \pm \frac{\sqrt{6}}{2}$