# What is the axis of symmetry and vertex for the graph y = 3x^2 - 7x - 8?

May 4, 2017

The axis of symmetry is $x = \frac{7}{6}$ and the vertex $\left(\frac{7}{6} , - \frac{145}{12}\right)$

#### Explanation:

Given a quadratic equation representing a parabola in the form:

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

we can convert to vertex form by completing the square:

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

$\textcolor{w h i t e}{y} = a {\left(x - \frac{- b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

$\textcolor{w h i t e}{y} = a {\left(x - h\right)}^{2} + k$

with vertex $\left(h , k\right) = \left(- \frac{b}{2 a} , c - {b}^{2} / \left(4 a\right)\right)$.

The axis of symmetry is the vertical line $x = - \frac{b}{2 a}$.

In the given example, we have:

$y = 3 {x}^{2} - 7 x - 8$

$\textcolor{w h i t e}{y} = 3 {\left(x - \frac{7}{6}\right)}^{2} - \left(8 + \frac{49}{12}\right)$

$\textcolor{w h i t e}{y} = 3 {\left(x - \frac{7}{6}\right)}^{2} - \frac{145}{12}$

So the axis of symmetry is $x = \frac{7}{6}$ and the vertex $\left(\frac{7}{6} , - \frac{145}{12}\right)$

graph{(y-(3x^2-7x-8))(4(x-7/6)^2+(y+145/12)^2-0.01)(x-7/6) = 0 [-5.1, 5.1, -13.2, 1.2]}