# How do you graph and label the vertex and axis of symmetry of y=2/3(x-2)^2-5/3?

Jul 20, 2017

Vertex$\to \left(x , y\right) = \left(2 , - \frac{5}{3}\right)$

Axis of symmetry $\to {x}_{\text{vertex}} \to x = 2$

${y}_{\text{intercept}} \to \left(x , y\right) = \left(0 , 1\right)$

x_("intercept")=2+-sqrt(10)/3 larr" Exact answer"
${x}_{\text{intercept}} = 3.58 \mathmr{and} 0.42$ to 2 decimal places (2dp)

#### Explanation:

This is the vertex form of a quadratic from which, with a small adjustment, you can directly read off the vertex.

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

Vertex form $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k + c$

Where $a \times {\left[\frac{b}{2 a}\right]}^{2} + k = 0$

In this case $k + c = - \frac{5}{3}$

$k$ is needed as changing $y = a {x}^{2} + b x + c$ into this form introduces an error. So the addition of $k$ acts as a correction factor.

Given:$y = \frac{2}{3} {\left(x - 2\right)}^{2} - \frac{5}{3}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the vertex}}$

x_("vertex")=(-1)xxb/(2a)" "=" "(-1)xx(-2)=+2

y_("vertex")=k+c" "=" "=-5/3

Vertex$\to \left(x , y\right) = \left(2 , - \frac{5}{3}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine y-intercept}}$

Set $x = 0$

$y = \frac{2}{3} {\left(0 - 2\right)}^{2} - \frac{5}{3}$

${y}_{\text{intercept}} = \frac{8}{3} - \frac{5}{3} = 1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine x-intercept}}$

Set $y = 0 = \frac{2}{3} {\left(x - 2\right)}^{2} - \frac{5}{3}$

Add $\frac{5}{3}$ to both sides

$\frac{5}{3} = \frac{2}{3} {\left(x - 2\right)}^{2}$

Multiply both sides by $\frac{3}{2}$

$\frac{15}{6} = {\left(x - 2\right)}^{2}$

Square root both sides

$\pm \sqrt{\frac{15}{6}} = x - 2$

x_("intercept")=2+-sqrt(15/6) larr" Exact answer"

Rounding decimals is never precise so the approximate answer is:

${x}_{\text{intercept}} = 3.58 \mathmr{and} 0.42$ to 2 decimal places (2dp) 