# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y = -3x^2 + 12x - 8?

Jun 13, 2015

Convert the equation into vertex form. Use the vertex and parabolic form to determine the axis of symmetry. Set one variable to $0$ to determine the intercept of the other.

#### Explanation:

Conversion to Vertex Form
$\textcolor{w h i t e}{\text{XXXX}}$ vertex form is $y = m {\left(x - a\right)}^{2} + b$ with vertex at $\left(a , b\right)$
$y = - 3 {x}^{2} + 12 x - 8$
$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$ $y = \left(- 3\right) \left({x}^{2} - 4 x\right) - 8$
$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$ $y = \left(- 3\right) \left({x}^{2} - 4 x + 4\right) - 8 + 12$
$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$ $y = \left(- 3\right) {\left(x - 2\right)}^{2} + 4$
The vertex is at $\left(2 , 4\right)$

Axis of Symmetry
The equation is a parabola in standard position (vertical axis) opening downward. So the axis of symmetry is a vertical line running through the vertex:
$\textcolor{w h i t e}{\text{XXXX}}$ $x = 2$.

Intercepts
$\textcolor{w h i t e}{\text{XXXX}}$ y-intercept
Set $x = 0$ in equation to determine the y-intercept:
$y = 3 {\left(0\right)}^{2} + 12 \left(0\right) - 8$
$y = - 8$
$\textcolor{w h i t e}{\text{XXXX}}$ the y-intercept is $\left(- 8\right)$

$\textcolor{w h i t e}{\text{XXXX}}$ x-intercept
Set $y = 0$ and solve for $x$ to determine the x-intercepts
$0 = 3 {x}^{2} + 12 x - 8$
Use the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
to determine x-intercepts

$x \text{ intercepts} = \frac{- 12 \pm \sqrt{144 + 96}}{6}$

$\textcolor{w h i t e}{\text{XXXX}}$ $= - 2 \pm \frac{\sqrt{240}}{6}$

$\textcolor{w h i t e}{\text{XXXX}}$ $= - 2 \pm \frac{2 \sqrt{15}}{3}$