Question

What is the the vertex of `y = (x -3)^2-9x+5`?

Answer 1

Vertex at : `(7 1/2,-42 1/4)`

Explanation:

Given
`color(white)("XXX")y=(x-3)^2-9x+5`

Expanding:
`color(white)("XXX")y=x^2-6x+9-9x+5`

`color(white)("XXX")y=x^2-15x+14`

We can proceed from here in 2 ways:

• by converting this into vertex form through "completing the square" method
• using the axis of symmetry (below)

Using the axis of symmetry
Factoring we have
`color(white)("XXX")y=(x-1)(x-14)`
which implies `y=0` (the X-axis) when `x=1` and when `x=14`

The axis of symmetry passes through the midpoint between the zeros
i.e. the axis of symmetry is `x=(1+14)/2= 15/2`

Note that the axis of symmetry also passes through the vertex;
so we can solve the original equation (or more easily our factored version) for the value of `y` where the equation and the axis of symmetry intersect:
`color(white)("XXX")y=(x-1)(x-14)` for `x=15/2`

`color(white)("XXX")rarr y=(15/2-1)(15/2-14)=13/2 * (-13/2))=-169/4`

So the vertex is at `(15/2,-169/4)=(7 1/2,-42 1/4)`

We can verify this result with a graph of the original equation:

graph{(x-3)^2-9x+5 [-0.016, 14.034, -45.34, -38.32]}