Question

What is the vertex and the equation of the axis of symmetry graph of `y=x^2-6x-7`?

Answer 1

The vertex is at `(3, -16)` and the axis of symmetry is `x=3`.

Explanation:

First, the EASY WAY to do this problem. For ANY quadratic equation in standard form

`y=ax^2+bx+c`

the vertex is located at `(-b/(2a),c-b^2/(4a))`.

In this case `a=1`, `b=-6`, and `c=-7`, so the vertex is at

`(-(-6)/(2*1),-7-(-6)^2/(4*1))=(3, -16)`.

But suppose you didn't know these formulae. Then the easiest way to get the vertex information is to convert the standard form quadratic expression into the vertex form `y=a(x-k)^2+h` by completing the square. The vertex will be at `(k, h)`.

`y=x^2-6x-7=x^2-6x+9-16=(x-3)^2-16`.

Again we see that the vertex is at `(3,-16)`.

The axis of symmetry for a parabola is always the vertical line containing the vertex (`x=k`), or in this case `x=3`.

graph{x^2-6x-7 [-10, 10, -20, 5]}

Answer 2

A different approach:

Axis of symmetry `->x=3`

Vertex `->(x,y)=(3,-16)`

Explanation:

Given: `y=x^2color(red)(-6)x-7`

What I am about to do is part of the process of completing the square.

`y=a(x+color(red)(b)/(2a))^2+k+c`

In this case `a=+1` so we ignore it.

Note that `color(red)(b=-6)`

`x_("vertex") = x_("axis of symmetry")=(-1/2)xxcolor(red)(b)`

`color(white)("dddddddddddddddddddd") (-1/2)color(red)(xx(-6))=+3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for `x=+3`

`y=x^2-6x-7color(white)("dddd")->color(white)("dddd")y=3^2-6(3)-7`

`color(white)("d"dddddddddddddddd.)->color(white)("dddd")y=-16`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Axis of symmetry `->x=3`

Vertex `->(x,y)=(3,-16)`