Where is tge vertex of the graph of y=-x^2

1 answer:

4 0

$\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
y=-x^2\implies y=-1(x-\stackrel{h}{0})^2+\stackrel{k}{0}$

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5 0

3 years ago

The base of a solid in the region bounded by the two parabolas y2 = 8x and x2 = 8y. Cross sections of the solid perpendicular to

Usimov [2.4K]

The two curves intersect at two points, (0, 0) and (8, 8):

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