Is this relationship linear, exponential, or neither?

Mathematics

1 answer:

Ratling [72]3 years ago

8 0

Answer:

It closely approximates an exponential function.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Now, assuming is a vertical parabola, so the squared variable is the "x", let's do some checking.

$\bf \qquad \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\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
vertex
\begin{cases}
h=3\\
k=5
\end{cases}\implies y=a(x-3)^2+5
\\\\\\
\textit{we also know that }
\begin{cases}
y=6\\
x=-1
\end{cases}\implies 6=a(-1-3)^2+5
\\\\\\
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