# What is the axis of symmetry and vertex for the graph y = –2x^2 – 32x – 126?

May 12, 2018

3 solution approaches

Vertex $\to \left(x , y\right) = \left(- 8 , 2\right)$

Axis of symmetry$\to x = - 8$

#### Explanation:

3 general conceptual options.

1: Determine the x-intercepts and the vertex is $\frac{1}{2}$ way between. Then use substitution to determine Vertex.

2: Complete the square and almost directly read of the vertex coordinates.

3: Start the 1st step of completing the square and use that to determine ${x}_{\text{vertex}}$. Then by substitution determine ${y}_{\text{vertex}}$
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Given: $y = - 2 {x}^{2} - 32 x - 126$

$\textcolor{b l u e}{\text{Option 1:}}$

Try to factorise $\to - 2 \left({x}^{2} + 16 x + 63\right) = 0$

Note that $9 \times 7 = 63 \mathmr{and} 9 + 7 = 16$

$- 2 \left(x + 7\right) \left(x + 9\right) = 0$

$x = - 7 \mathmr{and} x = - 9$

${x}_{\text{vertex}} = \frac{- 16}{2} = - 8$
By substitution you can determine ${y}_{\text{vertex}}$

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$\textcolor{b l u e}{\text{Option 2:}}$

Given: $y = - 2 {x}^{2} - 32 x - 126$

$y = - 2 \left({x}^{2} + 16 x\right) + k - 126 \leftarrow \text{ At this stage } k = 0$

Halve the 16, remove the $x$ from $16 x$ and move the squared.

$y = - 2 {\left(x + 8\right)}^{2} + k - 126 \leftarrow \text{ "k" now has a value}$

Set $- 2 {\left(8\right)}^{2} + k = 0 \implies k = 128$

$y = - 2 {\left(x + 8\right)}^{2} + 128 - 126$

$y = 2 {\left(x \textcolor{red}{+ 8}\right)}^{2} \textcolor{g r e e n}{+ 2}$

${x}_{\text{vertex}} = \left(- 1\right) \times \textcolor{red}{8} = \textcolor{m a \ge n t a}{- 8}$
Vertex $\to \left(x , y\right) = \left(\textcolor{m a \ge n t a}{- 8} , \textcolor{g r e e n}{2}\right)$
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$\textcolor{b l u e}{\text{Option 3:}}$
Given: $y = - 2 {x}^{2} - 32 x - 126$

$y = - 2 \left({x}^{2} + 16 x\right) + k - 126$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times 16 = - 8$

By substitution determine ${y}_{\text{vertex}}$