Question

How do you find the axis of symmetry and vertex point of the function: `y=4x^2-4x-15`?

Answer 1

Vertex: `(1/2,-16)`
Axis of symmetry: `x=1/2`

Explanation:

Given
`color(white)("XXX")y=4x^2-4x-15`

Part 1: The Vertex
Convert into vertex form (`g(x)=m(x-a)^2+b` with vertex at `(a,b)`)

`color(white)("XXX")`Extract the `m`
`color(white)("XXX")y)=4(x^2-1x) - 15`

`color(white)("XXX")`Complete the square
`color(white)("XXX")y= 4(x^2-1x+(1/2)^2)-15 - 4(1/2)^2)`

`color(white)("XXX")`Re-write as a squared binomial of form `(x-a)^2` and simply
`color(white)("XXX")y=4(x-(1/2))^2+(-16)`

This is in vertex form with the vertex at `(1/2,-16)`

Part 2: The Axis of Symmetry
An parabolic equation in the form:
`color(white)("XXX")y=ax^2+bx+c`
has a vertical axis (i.e. `x=c` for some constant `c`) through the vertex.

Since the given `y=)4x^2+4x-15` is of this form and the vertex is at `(x,y)=(1/2,-16)`
the axis of symmetry is
`color(white)("XXX")x=(1/2)`
graph{4x^2-4x-15 [-5.45, 14.56, -18.64, -8.64]}