3 years ago

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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KATRIN_1 [288]

Answer:

See Explanation

Step-by-step explanation:

Given

$Ratio=12:40$

Required

Determine the actual distance between x and y

The question is incomplete, as the scale distance between x and y is not given.

To solve this, we make assumptions.

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So, we have:

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Express as fraction

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Make Actual, the subject

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Readme [11.4K]

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