Question

Which of the following equations is a right-end behavior model for f(x)=−4x2+e−x ? A. y=−4x2 B. y=e−x C. y=ex D. y=4x E. y=−e−x

Answer 1

The right-end behavior model for
`\(f(x) = -4x^2 + e^(-x)\) is \(y = -4x^2\)`. Therefore, the correct answer is A. y=−4x2.

Step-by-step explanation:

In the given function
`\(f(x) = -4x^2 + e^(-x)\), as \(x\)`approaches positive infinity, the term
`\(e^(-x)\)` becomes negligible compared to the dominant term
`\(-4x^2\)`. This is because the exponential function
`\(e^(-x)\)` approaches zero faster than any power of
`\(x\) as \(x\)`becomes large. Therefore, in the right-end behavior, the function is effectively
`\(f(x) \approx -4x^2\),` leading to the right-end behavior model
`\(y = -4x^2\)` (Option A).

To understand this mathematically, consider taking the limit as
`\(x\)` approaches positive infinity:

`\[ \lim_{{x \to \infty}} (-4x^2 + e^(-x)) = \lim_{{x \to \infty}} -4x^2 + \lim_{{x \to \infty}} e^(-x) = -\infty + 0 = -\infty. \]`

This indicates that the exponential term
`\(e^(-x)\)` becomes negligible, and the leading term
`\(-4x^2\)` dominates the function's behavior. Therefore, the right-end behavior model is
`\(y = -4x^2\).`

In summary, the right-end behavior of the given function is primarily determined by the quadratic term, and as
`\(x\)` approaches positive infinity, the exponential term becomes insignificant. This leads to the conclusion that the right-end behavior model is
`\(y = -4x^2\)` .

Therefore, the correct answer is A. y=−4x2.