Question

What is the axis of symmetry and vertex for the graph `f(x)=x^2+2x-8`?

Answer 1

Vertex`" "->" "(x,y)" "->" "(-1,-9)`

Axis of symmetry`" " = " "x_("vertex")=-1`

Explanation:

The method I am about to use is the beginning part of completing the square.

Given:`" "f(x)=x^2+color(red)(2)x-8`

Compare to standard form of `ax^2+bx+c`

I can rewrite this as:`" " a(x^2+color(red)(b/a)x)+c`

I then apply: `" "(-1/2)xx color(red)(b/a) = x_("vertex")`

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`color(blue)("To determine "x_("vertex")`

In your case `a=1" and "b=2` so we have

`color(blue)(x_("vertex")=(-1/2)xx color(red)(2/1) =-1)`

Fast, isn't it!
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`color(blue)("Axis of symmetry = "x_("vertex")=-1`

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`color(blue)("To determine "y_("vertex")`

Substitute `color(blue)(x=-1)` in the original equation

`color(brown)(y_("vertex")=color(blue)((-1))^2+2color(blue)((-1))-8`

`color(blue)(y_("vertex")=1-2-8=-9`
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