# What are the important points needed to graph y=3(x+3)^2 - 3?

Mar 20, 2017

${y}_{\text{intercept}} = 24$
${x}_{\text{intercepts}} = - 4 \mathmr{and} - 2$
Vertex $\to \left(x , y\right) = \left(- 3 , - 3\right)$

#### Explanation:

This is the vertex form of the standardised$\text{ } y = a {x}^{2} + b x + c$
Multiplying out the brackets the we have $\text{ } y = 3 {x}^{2} + 18 x + 24$

As the $3 {x}^{2}$ is positive the graph is of general shape $\cup$

$\textcolor{b l u e}{\text{Determine the vertex using the questions equation format}}$

Given:$\text{ } y = 3 {\left(x \textcolor{red}{+ 3}\right)}^{2} \textcolor{g r e e n}{- 3}$

${x}_{\text{vertex}} = \left(- 1\right) \times \textcolor{red}{\left(+ 3\right)} = - 3$
${y}_{\text{vertex}} = \textcolor{g r e e n}{- 3}$

Vertex$\to \left(x , y\right) = \left(- 3 , - 3\right)$
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$\textcolor{b l u e}{\text{Determine the x-intercepts}}$

set $y = 0$

$\implies 0 = 3 {\left(x + 3\right)}^{2} - 3$

$3 = 3 {\left(x + 3\right)}^{2}$

${\left(x + 3\right)}^{2} = 1$

x+3=+-sqrt((1)

$x = - 3 \pm 1$

$x = - 4 \mathmr{and} - 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine y-intercept}}$

Set $x = 0$

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

$y = 27 - 3 = 24$

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$\textcolor{b l u e}{\text{Foot note}}$

Using the format $y = 3 {x}^{2} + 18 x + 24$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\frac{18}{3}\right) = - 3$