Question

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Answer 1

Arrange it to slope intercept form

Answer 2

\[2x^2 + 32y = 32 \\ 32y=-2x^2+32 \\ y=-\frac{1}{16}x^2+1\]
We'll use \(\huge x=-\frac{b}{2a}\) to find the vertex, since 'b' is zero, The point x of vertex is zero.
When x=0,
y=1

So vertex is (0,1)

Answer 3

How do I get the focus?

Answer 4

Write it \[y=-\frac{1}{16}x^2\]
Then use the form:
\[x^2=(4p)y\]
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using our original equation
\[y=-\frac{1}{16}x^2\]
\[-16y=x^2\]
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So, 4p=-16
p=-4

Then add 1 for p because
\[y=-\frac{1}{16}x^2(+1)\]
p=-3
Focus will be in the form (0,p)