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

May 18, 2017

#### Explanation:

An equation of the form $y = a {\left(x - h\right)}^{2} + k$ has $x - h = 0$ as axis of symmetry and $\left(h , k\right)$ as vertex. Such an equation is a vertical parabola and if $a$ is positive, it open up and if $a$ is negative it opens down.

Hence in $y = \frac{2}{3} {\left(x - 2\right)}^{2} - \frac{5}{3}$, we have a parabola whose vertex is $\left(2 , \frac{5}{3}\right)$, axis of symmetry is $x - 2 = 0$ and it opens up. As it opens up, we have the minima appearing at vertex i.e. at $\left(2 , - \frac{5}{3}\right)$.

To draw the parabola select a few points around $x = 2$ i.e. on both sides of axis of symmetry. Here I select values of $x$ as $\left\{- 7 , - 4 , - 1 , 2 , 5 , 8 , 11\right\}$ and find the corresponding value of $y$ given by

$y = \frac{2}{3} {\left(x - 2\right)}^{2} - \frac{5}{3}$ and I get the points through which parabola passes as $\left(- 7 , 52 \frac{1}{3}\right) , \left(- 4 , 22 \frac{1}{3}\right) , \left(- 1 , 4 \frac{1}{3}\right) , \left(2 , - 1 \frac{2}{3}\right) , \left(5 , 4 \frac{1}{3}\right) , \left(8 , 22 \frac{1}{3}\right) , \left(11 , 52 \frac{1}{3}\right)$. Observe that I have selected points at an interval of $3$ so that they get cancelled out as we have $a = \frac{2}{3}$.

Joining these points gives us the following graph and we have marked axis of symmetry and vertex too.

graph{(2/3(x-2)^2-5/3-y)(x-2)((x-2)^2+(y+5/3)^2-0.05)=0 [-18.3, 21.7, -3.04, 16.96]}