# How do you find the axis of symmetry, and the maximum or minimum value of the function y=-3x^2-12x-3?

Apr 30, 2018

$x = - 2 , \text{ max value } = 9$

#### Explanation:

"given a quadratic in standard form ";ax^2+bx+c;a!=0

$\text{then the x-coordinate of the vertex which is also the}$
$\text{axis of symmetry can be found using}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$\text{here } a = - 3 , b = - 12 , c = - 3$

$\Rightarrow {x}_{\text{vertex}} = - \frac{- 12}{-} 6 = - 2$

$\Rightarrow \text{axis of symmetry is } x = - 2$

$\text{the vertex lies on the axis of symmetry thus substituting}$
$\text{x = - 2 into the equation gives y-coordinate}$

$\Rightarrow {y}_{\text{vertex}} = - 3 {\left(- 2\right)}^{2} - 12 \left(- 2\right) - 3 = 9$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 2 , 9\right)$

$\text{to determine if vertex is maximum/minimum}$

• " if "a>0" then minimum turning point"

• " if "a<0" then maximum turning point"

$\text{here } a = - 3 < 0$

$\Rightarrow - 3 {x}^{2} - 12 x - 3 \text{ has a maximum at } \left(- 2 , \textcolor{red}{9}\right)$

$\text{with maximum value } = 9$
graph{-3x^2-12x-3 [-20, 20, -10, 10]}