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vesna_86 [32]
2 years ago
8

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

Mathematics
1 answer:
Reika [66]2 years ago
8 0

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.

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

The given function is:

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

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

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

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

\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.

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

Question c:

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

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

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)

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

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

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

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