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RoseWind [281]
3 years ago
14

Which of the following exponential functions goes thru points (1,6) and (2,12)

Mathematics
1 answer:
Alexandra [31]3 years ago
7 0

Answer:

Noting that e^b = 6/a

Step-by-step explanation:

Use the general expo function y = ae^(bx).

Subbing 6 for y and 1 for x, we get  6 = ae^(b), or   e^b = 6/a.

Subbing 12 for y and 2 for x, we get   12 = ae^(2b), or 12/a = (e^b)²

Now let's find the value of the coefficient a.  Noting that e^b = 6/a, rewrite

12/a = (e^b)² as 12/a = (6/a)².

Dividing both sides by 6/a, we get 2 = 6/a, or a = 3.

Again Noting that e^b = 6/a, e^b = 6/3, or e^b = 2.

Taking the natural log of both sides, we get b = ln 2.

Then our y = ae^(bx) becomes:

                y = 3e^(ln 2·x), or   y = 3·2^x

Note:  next time, please share the answer choices.  Thank you.

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serg [7]
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now, the "p" distance is 2, however, the directrix is up, the focus point is below it, the parabola opens towards the focus point, thus, the parabola is opening downwards, and the squared variable is the "x"

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now, let's plug all those fellows in then

\bf \begin{array}{llll}
(x-{{ h}})^2=4{{ p}}(y-{{ k}})\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-----------------------------\\\\

\begin{cases}
h=3\\
k=3\\
p=-2
\end{cases}\implies (x-3)^2=4(-2)(y-3)\implies (x-3)^2=-8(y-3)
\\\\\\
-\cfrac{(x-3)^2}{8}=y-3\implies \boxed{-\cfrac{1}{8}(x-3)^2+3=y}

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Answer:

$16, sorry if its wrong (im pretty sure i'm right tho)

6 0
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Read 2 more answers
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