Question

Given f of x is equal to the quantity 4x minus 7 end quantity divided by the quantity 8x plus 8 end quantity, what is the end behavior of the function?

As x → -∞, f(x) → 0.5; as x → ∞, f(x) → 0.5.
As x → -∞, f(x) → -0.5; as x → ∞, f(x) → -0.5.
As x → -∞, f(x) → 0.73; as x → ∞, f(x) → 0.73.
As x → -∞, f(x) → -0.73; as x → ∞, f(x) → -0.73.

Answer 1

To analyze the end behavior of the function

\[
f(x) = \frac{4x - 7}{8x + 8},
\]

we can look at the behavior of the function as \( x \to \infty \) and \( x \to -\infty \).

1. **As \( x \to \infty \)**:
- To simplify the function, we can divide the numerator and the denominator by \( x \):

\[
f(x) = \frac{4 - \frac{7}{x}}{8 + \frac{8}{x}}.
\]

- As \( x \) approaches infinity, \( \frac{7}{x} \) approaches 0 and \( \frac{8}{x} \) also approaches 0. Thus the function simplifies to:

\[
f(x) \to \frac{4 - 0}{8 + 0} = \frac{4}{8} = \frac{1}{2}.
\]

2. **As \( x \to -\infty \)**:
- We apply the same division by \( x \):

\[
f(x) = \frac{4 - \frac{7}{x}}{8 + \frac{8}{x}}.
\]

- Again, as \( x \) approaches negative infinity, \( \frac{7}{x} \) approaches 0 and \( \frac{8}{x} \) approaches 0. Thus the function simplifies to:

\[
f(x) \to \frac{4 - 0}{8 + 0} = \frac{4}{8} = \frac{1}{2}.
\]

Since we find that both the limits as \( x \to \infty \) and \( x \to -\infty \) approach \( \frac{1}{2} \), we conclude:

- As \( x \to -\infty \), \( f(x) \to 0.5 \);
- As \( x \to \infty \), \( f(x) \to 0.5 \).

Therefore, the correct choice is:

**As \( x \to -\infty, f(x) \to 0.5; \text{ as } x \to \infty, f(x) \to 0.5.**