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

Oct 5, 2017

Vertex at : $\left(7 \frac{1}{2} , - 42 \frac{1}{4}\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - 3\right)}^{2} - 9 x + 5$

Expanding:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 6 x + 9 - 9 x + 5$

$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 15 x + 14$

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
$\textcolor{w h i t e}{\text{XXX}} y = \left(x - 1\right) \left(x - 14\right)$
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 $x = \frac{1 + 14}{2} = \frac{15}{2}$

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:
$\textcolor{w h i t e}{\text{XXX}} y = \left(x - 1\right) \left(x - 14\right)$ for $x = \frac{15}{2}$

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

So the vertex is at $\left(\frac{15}{2} , - \frac{169}{4}\right) = \left(7 \frac{1}{2} , - 42 \frac{1}{4}\right)$

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]}