# What is the vertex of  y=5(x/3-15)^2-4 ?

Feb 25, 2016

Vertex $\left(45 , - 4\right)$

#### Explanation:

There are a couple ways of doing this; perhaps the most obvious is to convert the given equation into standard vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - a\right)}^{2} + b$ with its vertex at $\left(a , b\right)$

$y = 5 {\left(\frac{x}{3} - 15\right)}^{2} - 4$

$\rightarrow y = 5 {\left(\frac{x - 45}{3}\right)}^{2} - 4$

$\rightarrow \frac{5}{9} {\left(x - 45\right)}^{2} + \left(- 4\right)$
$\textcolor{w h i t e}{\text{XXX}}$which is the vertex form with vertex at $\left(45 , - 4\right)$

Alternately think of replacing $\hat{x} = \frac{x}{3}$ and the given equation is in vertex form for $\left(\hat{x} , y\right) = \left(15 , - 4\right)$
and since $x = 3 \cdot \hat{x}$ the vertex using $x$ is $\left(x , y\right) = \left(3 \times 15 , - 4\right)$
graph{5(x/3-15)^2-4 [35.37, 55.37, -6.36, 3.64]}