# How do you find the axis of symmetry, and the maximum or minimum value of the function y=x^2-8x+3?

Jan 30, 2016

This form has a vertical axis of symmetry through the vertex, opening upward. Find the vertex to get the minimum (the $y$ coordinate) and the axis of symmetry ($x =$ the $x$ coordinate)

#### Explanation:

Any parabolic equation in the form:
$\textcolor{w h i t e}{\text{XXX}} y = p {x}^{2} + q x + r$
has a vertical axis of symmetry and
$\textcolor{w h i t e}{\text{XXX}}$if $p > 0$ opens upward;
$\textcolor{w h i t e}{\text{XXX}}$if $p < 0$ opens downward.

Since
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 8 x + 3$
has an implied value of $p = + 1$
it opens upward $\Rightarrow$ its vertex is its minimum.

Converting $y = {x}^{2} - 8 x + 3$ into vertex form:

Complete the square:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 8 x \textcolor{g r e e n}{+ 16} + 3 \textcolor{g r e e n}{- 16}$

$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - 4\right)}^{2} - 13$
which is the vertex form with the vertex at $\left(4 , - 13\right)$

The minimum value is $\left(- 13\right)$
and the axis of symmetry is $x = 4$.

For verification purposes here is the graph of $y = {x}^{2} - 8 x + 3$
graph{x^2-8x+3 [-15.56, 16.48, -15.01, 1.01]}