# What is the axis of symmetry and vertex for the graph f(x)=-3x^2+6x+12?

Jun 16, 2017

Axis of symmetry is $x = 1$, vertex is at $\left(1 , 15\right)$.

#### Explanation:

$f \left(x\right) = - 3 {x}^{2} + 6 x + 12 = - 3 \left({x}^{2} - 2 x\right) + 12 = - 3 \left({x}^{2} - 2 x + 1\right) + 3 + 12$

$= - 3 {\left(x - 1\right)}^{2} + 15$ . Comparing with standard vertex form of equation f(x)= a(x-h)^2+k ; (h,k) being vertex.

Here $h = 1 , k = 15$ . So vertex is at $\left(1 , 15\right)$.

Axis of symmetry is $x = 1$

graph{-3x^2+6x+12 [-40, 40, -20, 20]} [Ans]

Jun 16, 2017

$x = 1 , \text{vertex } = \left(1 , 15\right)$

#### Explanation:

$\text{for a parabola in standard form } y = a {x}^{2} + b x + c$

"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)

$y = - 3 {x}^{2} + 6 x + 12 \text{ is in standard form}$

$\text{with " a=-3, b=6" and } c = 12$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{6}{- 6} = 1$

$\text{substitute this value into function for y-coordinate}$

${y}_{\textcolor{red}{\text{vertex}}} = - 3 + 6 + 12 = 15$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , 15\right)$

$\text{since " a<0" then graph has a maximum } \bigcap$

$\text{the axis of symmetry passes through the vertex }$

$\Rightarrow x = 1 \text{ is equation of axis of symmetry}$
graph{(y+3x^2-6x-12)(y-1000x+1000)=0 [-40, 40, -20, 20]}