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LenaWriter [7]
3 years ago
12

1. The value -2 is a lower bound for the zeros of the function shown below. f(x)=4x^3-12x^2-x+15

Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
3 0
First is false i believe
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5.6%

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Studentka2010 [4]

Part A: Vertical asymptote is x=0

Part B: Domain is \left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)

Part C: Horizontal asymptote is y=4

Part D: Range is \left(-\infty \:,\:4\right)\cup \left(4,\:\infty \:\right)

Explanation:

Part A: We need to determine the vertical asymptote

The vertical asymptote of a function can be determined by equating the denominator equal to zero.

Thus, we have,

x=0

Hence, the vertical asymptote is x=0

Part B: We need to determine the domain

The domain of the function is the set of all independent x - values for which the function is real and well defined.

Let us take the denominator and equate to zero.

Hence, we have, x=0

Therefore, the function is undefined at the point x=0

Thus, the domain of the function is \left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)

Part C: We need to determine the horizontal asymptote

The horizontal asymptote of the function can be determined by dividing the leading coefficient of the numerator by leading coefficient of the denominator.

Thus, we have, y=4

Hence, the horizontal asymptote of the function is y=4

Part D: We need to determine the range

The range of the function is the set of all dependent y -values of the function.

In other words, the range of the function can be determined by substituting the values for x.

Thus, we have,

\left(-\infty \:,\:4\right)\cup \left(4,\:\infty \:\right)

Therefore, the range of the function is \left(-\infty \:,\:4\right)\cup \left(4,\:\infty \:\right)

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