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baherus [9]
2 years ago
5

The graph of this function is shifted 7 units right and shifted 5 units up.

Mathematics
1 answer:
mojhsa [17]2 years ago
6 0
<h2><u>Graphing</u></h2>

<h3>The graph of this function is shifted 7 units right and shifted 5 units up.</h3>

<h3>f(x) = -2/3(x - 7)² - 5</h3>

  • True
  • False

<u>Answer:</u>

  • <u>False</u>

<u>Explanation:</u>

  • Why false? As I noticed, in the equation f(x) = -2/3(x - 7)² - 5, the vertex is (7, -5). The value of a, which is -2/3, is negative. So the graph opens downwards. The graph shifts 7 units to the right, which is correct, but graphing 5 units up is wrong. Why? 5 is negative, so it should be shifted 5 units down.

Wxndy~~

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Answer:

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Step-by-step explanation:

By the double angle identity of sines:

\sin(2\, x) = 2\, \sin x \cdot \cos x.

Rewrite the original equation with this identity:

2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0.

Note, that 2\, (2\, \sin x \cdot \cos x) and (-2\, \sin x) share the common factor (2\, \sin x). On the other hand, 2\sqrt{3}\, \cos x and (-\sqrt{3}) share the common factor \sqrt[3}. Combine these terms pairwise using the two common factors:

(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0.

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This equation holds as long as either \left(2\, \sin x + \sqrt{3}\right) or (2\, \cos x - 1) is zero. Let k be an integer. Accordingly:

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Answer:

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Step-by-step explanation

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<em>Sorry if this is wrong.</em>

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