# What is the axis of symmetry and vertex for the graph y=x^2-3x+8?

Jan 3, 2018

Vertex $\left(\frac{3}{2} , \frac{23}{4}\right)$
Axis of symmetry: $x = \frac{3}{2}$

#### Explanation:

Given a quadratic of the form $y = a {x}^{2} + b x + c$ the vertex, $\left(h , k\right)$ is of the form $h = - \frac{b}{2 a}$ and $k$ is found by substituting $h$.

$y = {x}^{2} - 3 x + 8$ gives $h = - \frac{- 3}{2 \cdot 1} = \frac{3}{2}$.

To find $k$ we substitute this value back in:

$k = {\left(\frac{3}{2}\right)}^{2} - 3 \left(\frac{3}{2}\right) + 8 = \frac{9}{4} - \frac{9}{2} + 8 = \frac{23}{4}$.

So the vertex is $\left(\frac{3}{2} , \frac{23}{4}\right)$.

The axis of symmetry is the vertical line through the vertex, so in this case it is $x = \frac{3}{2}$.