# What is the axis of symmetry and vertex for the graph y = 9x^2 - 27x + 20?

Jul 13, 2018

The axis of symmetry is $x = \frac{3}{2}$.

The vertex is $\left(\frac{3}{2} , - \frac{1}{4}\right)$.

#### Explanation:

Given:

$y = 9 {x}^{2} - 27 x + 20$ is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

where:

$a = 9$, $b = 027$, $c = 20$

The formula for the axis of symmetry is:

$x = \frac{- b}{2 a}$

$x = \frac{- \left(- 27\right)}{2 \cdot 9}$

$x = \frac{27}{18}$

Reduce by dividing the numerator and denominator by $9$.

$x = \frac{27 \div 9}{18 \div 9}$

$x = \frac{3}{2}$

The axis of symmetry is $x = \frac{3}{2}$. This is also the x-coordinate of the vertex.

To find the y-coordinate of the vertex, substitute $\frac{3}{2}$ for $x$ in the equation and solve for $y$.

$y = 9 {\left(\frac{3}{2}\right)}^{2} - 27 \left(\frac{3}{2}\right) + 20$

$y = 9 \left(\frac{9}{4}\right) - \frac{81}{2} + 20$

$y = \frac{81}{4} - \frac{81}{2} + 20$

The least common denominator is $4$. Multiply $\frac{81}{2}$ by $\frac{2}{2}$ and $20$ by $\frac{4}{4}$ to get equivalent fractions with $4$ as the denominator. Since $\frac{n}{n} = 1$, the numbers will change but the value of the fractions will remain the same.

$y = \frac{81}{4} - \left(\frac{81}{2} \times \frac{2}{2}\right) + \left(20 \times \frac{4}{4}\right)$

$y = \frac{81}{4} - \frac{162}{4} + \frac{80}{4}$

$y = \frac{81 - 162 + 80}{4}$

$y = - \frac{1}{4}$

The vertex is $\left(\frac{3}{2} , - \frac{1}{4}\right)$.

graph{y=9x^2-27x+20 [-10, 10, -5, 5]}