Given f of x is equal to the quantity 8x plus 1 end quantity divided by the quantity 2x minus 9 end quantity, what is the end behavior of the function?

Question

As x → -∞, f(x) → 9 ; as x → ∞, f(x) → 9.
As x → -∞, f(x) → -9; as x → ∞, f(x) → -9.
As x → -∞, f(x) → -4; as x → ∞, f(x) → -4.
As x → -∞, f(x) → 4; as x → ∞, f(x) → 4.

Answer 1

To analyze the end behavior of the function $f(x)=\frac{8x+1}{2x-9}$, we can find the limits as $x$ approaches $-\mathrm{\infty }$ and $+\mathrm{\infty }$.

1. As $x\to \mathrm{\infty }$: $$f(x)=\frac{8x+1}{2x-9}$$ We can divide the numerator and the denominator by $x$: $$=\frac{8+\frac{1}{x}}{2-\frac{9}{x}}$$ As $x\to \mathrm{\infty }$, $\frac{1}{x}\to 0$ and $\frac{9}{x}\to 0$: $$\underset{x\to \mathrm{\infty }}{lim}f(x)=\frac{8+0}{2-0}=\frac{8}{2}=4$$

2. As $x\to -\mathrm{\infty }$: Using the same approach: $$f(x)=\frac{8x+1}{2x-9}=\frac{8+\frac{1}{x}}{2-\frac{9}{x}}$$ Again, as $x\to -\mathrm{\infty }$, $\frac{1}{x}\to 0$ and $\frac{9}{x}\to 0$: $$\underset{x\to -\mathrm{\infty }}{lim}f(x)=\frac{8+0}{2-0}=\frac{8}{2}=4$$

Thus, the end behavior of the function is:

• As $x\to -\mathrm{\infty },f(x)\to 4$
• As $x\to \mathrm{\infty },f(x)\to 4$

Therefore, the correct answer is: As $x\to -\mathrm{\infty },f(x)\to 4;$ as ( x \to \infty, f(x) \to 4.