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

Jul 25, 2016

Vertex$\to \left(x , y\right) = \left(6 , 32\right)$

Axis of symmetry is: $x = 6$

#### Explanation:

Given:$\text{ } y = - {x}^{2} + 12 x - 4$
You can solve the traditional way or use a 'trick'

Just to give you an idea how useful the trick is:

By sight: $\textcolor{b r o w n}{\text{The axis of symmetry is } x = + 6}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine axis of symmetry and "x_("vertex"))

Consider the standard form of $y = a {x}^{2} + b x + c$

Write as: $y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

In your case $a = - 1$

So $\textcolor{b r o w n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{12}{- 1} = + 6}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

color(blue)("Determine "y_("vertex"))

Substitute $x = 6$ into original equation.

${y}_{\text{vertex")=-(6^2)+12(6)-4" "->" "y_("vertex}} = 32$

$\textcolor{w h i t e}{.}$

$\textcolor{m a \ge n t a}{\text{'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$
$\textcolor{m a \ge n t a}{\text{'~~~~~~~~~~~~Another method ~~~~~~~~~~~~~}}$
$\textcolor{m a \ge n t a}{\text{'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

$\textcolor{b l u e}{\text{Completing the square "color(brown)(larr" not much detail given}}$

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

$y = - {\left(x - 6\right)}^{2} - 4 + k$

$B u t - 36 + k = 0 \to k = 36$

$y = - {\left(x - 6\right)}^{2} + 32$

${x}_{\text{vertex}} \to \left(- 1\right) \times \left(- 6\right) = + 6$
${y}_{\text{vertex}} \to 32$