# What is the axis of symmetry and vertex for the graph y = -3x^2 + 12x + 4?

Jun 3, 2018

aos = 2

vertex = (2,16)

#### Explanation:

$y = - 3 {x}^{2} + 12 x + 4$

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

In the form $y = a {x}^{2} + b x + c$ you have:

$a = - 3$

$b = 12$

$c = 4$

Axis of symmetry (aos) is: $a o s = \frac{- b}{2 a} = \frac{- 12}{2 \cdot - 3} = 2$

Remember $y = f \left(x\right)$

Vertex is: $\left(a o s , f \left(a o s\right)\right) = \left(2 , f \left(2\right)\right)$:

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

$f \left(2\right) = - 3 {\left(2\right)}^{2} + 12 \cdot 2 + 4 = 16$

vertex $= \left(2 , 16\right)$

graph{-3x^2 + 12x + 4 [-16.71, 23.29, -1.6, 18.4]}

Jun 3, 2018

Vertex -

$\left(2 , 16\right)$

Axis of symmetry

$x = 2$

#### Explanation:

Given -

$y = - 3 {x}^{2} + 12 x + 4$

Vertex -

$x = \frac{- b}{2 a} = \frac{- 12}{2 \times - 3} = \frac{- 12}{- 6} = 2$

At x=2; y=-3(2^2)+12(2)+4

$y = - 12 + 24 + 4 = 16$

$\left(2 , 16\right)$

Axis of symmetry

$x = 2$