Question

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

Answer 1

3 solution approaches

Vertex `->(x,y)=(-8,2)`

Axis of symmetry`-> x=-8`

Explanation:

3 general conceptual options.

1: Determine the x-intercepts and the vertex is `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_("vertex")`. Then by substitution determine `y_("vertex")`
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Given: `y=-2x^2-32x-126`

`color(blue)("Option 1:")`

Try to factorise `-> -2(x^2+16x+63)=0`

Note that `9xx7=63 and 9+7=16`

`-2(x+7)(x+9)=0`

`x=-7 and x=-9`

`x_("vertex")=(-16)/2=-8`
By substitution you can determine `y_("vertex")`

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`color(blue)("Option 2:")`

Given: `y=-2x^2-32x-126`

`y=-2(x^2+16x)+k-126 larr" At this stage "k=0`

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

`y=-2(x+8)^2+k-126 larr" "k" now has a value"`

Set `-2(8)^2+k=0 => k = 128`

`y=-2(x+8)^2+128-126`

`y=2(xcolor(red)(+8))^2color(green)(+2)`

`x_("vertex")=(-1)xxcolor(red)(8)=color(magenta)(-8)`
Vertex `->(x,y)=(color(magenta)(-8),color(green)(2))`
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`color(blue)("Option 3:")`
Given: `y=-2x^2-32x-126`

`y=-2(x^2+16x)+k-126`

`x_("vertex")=(-1/2)xx16 = -8`

By substitution determine `y_("vertex")`