How do you find the axis of symmetry and vertex point of the function: y=4x^2-4x-15?

Oct 9, 2015

Vertex: $\left(\frac{1}{2} , - 16\right)$
Axis of symmetry: $x = \frac{1}{2}$

Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} y = 4 {x}^{2} - 4 x - 15$

Part 1: The Vertex
Convert into vertex form ($g \left(x\right) = m {\left(x - a\right)}^{2} + b$ with vertex at $\left(a , b\right)$)

$\textcolor{w h i t e}{\text{XXX}}$Extract the $m$
color(white)("XXX")y)=4(x^2-1x) - 15

$\textcolor{w h i t e}{\text{XXX}}$Complete the square
color(white)("XXX")y= 4(x^2-1x+(1/2)^2)-15 - 4(1/2)^2)

$\textcolor{w h i t e}{\text{XXX}}$Re-write as a squared binomial of form ${\left(x - a\right)}^{2}$ and simply
$\textcolor{w h i t e}{\text{XXX}} y = 4 {\left(x - \left(\frac{1}{2}\right)\right)}^{2} + \left(- 16\right)$

This is in vertex form with the vertex at $\left(\frac{1}{2} , - 16\right)$

Part 2: The Axis of Symmetry
An parabolic equation in the form:
$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
has a vertical axis (i.e. $x = c$ for some constant $c$) through the vertex.

Since the given y=)4x^2+4x-15 is of this form and the vertex is at $\left(x , y\right) = \left(\frac{1}{2} , - 16\right)$
the axis of symmetry is
$\textcolor{w h i t e}{\text{XXX}} x = \left(\frac{1}{2}\right)$
graph{4x^2-4x-15 [-5.45, 14.56, -18.64, -8.64]}