# How do you find the vertex and the intercepts for y = -3x^2 + 12x - 8?

Apr 13, 2017

Explained below, which is self explanatory.

#### Explanation:

$y = - 3 {x}^{2} + 12 x - 8$

$= - 3 \left({x}^{2} + 4 x\right) - 8$

$= - 3 \left({x}^{2} + 4 x + 4 - 4\right) - 8$

$= - 3 \left({x}^{2} + 4 x + 4\right) + 12 - 8$

$= - 3 {\left(x + 2\right)}^{2} + 4$, is the required vertex form.

This quadratic function represents a vertical parabola opening down.

The vertex is at $\left(- 2 , 4\right)$.

Axis of symmetry is $x = - 2$

$y$- intercept is given by $x = 0$. So it is $- 8$

$x$- intercept is given by $y = 0$.

$x$-intercepts would be given by the quadratic equation:

$- 3 {x}^{2} + 12 x - 8 = 0$,

or, $3 {x}^{2} - 12 x + 8 = 0$,

$x = \frac{12 \pm \sqrt{144 - 96}}{6} \text{ }$ using the quadratic formula.:

$x = \frac{12 \pm \sqrt{48}}{6}$

= $2 \pm \frac{\sqrt{48}}{6}$

$x = 3.15 \mathmr{and} x = 0.85 \text{ }$ (to 2 d.p.)