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Mathematics

1 answer:

denpristay [2]1 year ago

5 0

Answer:

$\huge\frac{ \sqrt{6} }{ \sqrt{5} } = \frac{ \sqrt{30} }{5}$

<h2>√30/5 is the right answer.</h2>

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For the function g(x)=3-8(1/4)^2-x

Reika [66]

Using function concepts, it is found that:

a) The y-intercept is y = 2.5.

b) The horizontal asymptote is x = 3.

c) The function is decreasing.

d) The domain is $(-\infty,\infty)$ and the range is $(-\infty,3)$ .
e) The graph is given at the end of the answer.

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The given function is:

$g(x) = 3 - 8\left(\frac{1}{4}\right)^{2-x}$

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Question a:

The y-intercept is g(0), thus:

$g(0) = 3 - 8\left(\frac{1}{4}\right)^{2-0} = 3 - 8\left(\frac{1}{4}\right)^{2} = 3 - \frac{8}{16} = 3 - 0.5 = 2.5$

The y-intercept is y = 2.5.

------------------------------------

Question b:

The horizontal asymptote is the limit of the function when x goes to infinity, if it exists.

$\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2+\infty} = 3 - 8\left(\frac{1}{4}\right)^{\infty} = 3 - 8\frac{1^{\infty}}{4^{\infty}} = 3 -0 = 3$

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$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2-\infty} = 3 - 8\left(\frac{1}{4}\right)^{-\infty} = 3 - 8\times 4^{\infty} = 3 - \infty = -\infty$

Thus, the horizontal asymptote is x = 3.

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Question c:

The limit of x going to infinity of the function is negative infinity, which means that the function is decreasing.

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Question d:

Exponential function has no restrictions in the domain, so it is all real values, that is $(-\infty,\infty)$ .
From the limits in item c, the range is: $(-\infty,3)$

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The sketching of the graph is given appended at the end of this answer.

A similar problem is given at brainly.com/question/16533631