Question

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

Answer 1

`y_("intercept")=+32`

Vertex`->(x,y)=(5-3)`

As `x_("vertex")=+5` and the graph is of shape `sub` then:
THERE IS NO Y INTERCEPT

Explanation:

Given:`" "x=3(y+3)^2+5`

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`color(blue)("What is normally considered as "x" has to be considered as "y)`
Thus INSTEAD OF VERTEX `(x,y)=(-3,5)` WE HAVE VERTEX `(x,y)=(5,-3)`
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Multiply out the brackets and we have:

`x=3y^2+18ycolor(red)(+32)`

This is a quadratic in y so and as the `y^2` term is positive it is of general shape: `sub`

Thus:

`color(red)(y_("intercept")=+32)`

Vertex`->(x,y)=(5,-3) larr" graph of shape" sub`