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galben [10]
3 years ago
5

The graph of g(x) is a transformation of the graph of f(x)=2x. Enter the equation for g(x) in the box. g(x) =

Mathematics
1 answer:
Lunna [17]3 years ago
4 0
<h3>Answer:  g(x) = 2^x - 3</h3>

The -3 is not in the exponent

Explanation:

The parent function is f(x) = 2^x. Plugging in x = 0 leads to y = 1. So the point (0,1) is on the f(x) curve. Going from (0,1) to (0,-2) is a vertical shift of 3 units downward. To represent this shift, we tack on a "-3" at the end of the f(x) function.

g(x) = f(x) - 3\\\\g(x) = 2^x - 3

You could look at other points as well, but I find working with x = 0 is easiest.

As a check, plugging x = 0 into g(x) leads to...

g(x) = 2^x - 3\\\\g(0) = 2^0 - 3\\\\g(0) = 1 - 3\\\\g(0) = -2

This confirms our answer.

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\sf n^{th} roots of a complex number is given by DeMoivre's formula.

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          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

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For k =2,

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For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

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