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

Jun 4, 2015

$y = f \left(x\right) = - 3 {x}^{2} - 12 x - 3 = - 3 \left({x}^{2} + 4 x + 1\right)$

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

This has its vertex when $x + 2 = 0$, that is when $x = - 2$.

$f \left(- 2\right) = - 3 \left({\left(- 2\right)}^{2} + 4 \left(- 2\right) + 1\right) = - 3 \left(4 - 8 + 1\right) = - 3 \times - 3 = 9$

So the vertex is at $\left(- 2 , 9\right)$ and the axis of symmetry is the line $x = - 2$

The $x$-axis intercepts are where $f \left(x\right) = 0$, which will happen when ${\left(x + 2\right)}^{2} - 3 = 0$, that is when

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

So

$x + 2 = \pm \sqrt{3}$

Subtract $2$ from both sides to get

$x = - 2 \pm \sqrt{3}$

The $y$-axis intercept is at $\left(0 , f \left(0\right)\right) = \left(0 , - 3\right)$

graph{-3x^2-12x-3 [-20, 20, -10, 10]}