# How do you find the vertex and intercepts for x=3(y+3)^2+5?

Mar 12, 2017

${y}_{\text{intercept}} = + 32$

Vertex$\to \left(x , y\right) = \left(5 - 3\right)$

As ${x}_{\text{vertex}} = + 5$ and the graph is of shape $\subset$ then:
THERE IS NO Y INTERCEPT

#### Explanation:

Given:$\text{ } x = 3 {\left(y + 3\right)}^{2} + 5$

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$\textcolor{b l u e}{\text{What is normally considered as "x" has to be considered as } y}$
Thus INSTEAD OF VERTEX $\left(x , y\right) = \left(- 3 , 5\right)$ WE HAVE VERTEX $\left(x , y\right) = \left(5 , - 3\right)$
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Multiply out the brackets and we have:

$x = 3 {y}^{2} + 18 y \textcolor{red}{+ 32}$

This is a quadratic in y so and as the ${y}^{2}$ term is positive it is of general shape: $\subset$

Thus:

$\textcolor{red}{{y}_{\text{intercept}} = + 32}$

Vertex$\to \left(x , y\right) = \left(5 , - 3\right) \leftarrow \text{ graph of shape} \subset$