for the function ( f(x)=2^{x}-1 ), find:\na. the ( y )-intercept.\n( (0, quad) )\nb. the behavior of the…

Explanation:

Step1: Find the y - intercept

The y - intercept occurs when (x = 0). Substitute (x=0) into (f(x)=2^{x}-1). (f(0)=2^{0}-1) Since (a^{0}=1) ((a\neq0)), then (2^{0}=1). So (f(0)=1 - 1=0).

Step2: Analyze the behavior as (x\to-\infty)

We know that the exponential function (y = a^{x}), when (0\lt a\lt1), (\lim_{x\to-\infty}a^{x}=\infty) and when (a > 1) (here (a = 2)), (\lim_{x\to-\infty}2^{x}=0). For (f(x)=2^{x}-1), (\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(2^{x}-1)) Using the limit rule (\lim_{x\to c}(u(x)-v(x))=\lim_{x\to c}u(x)-\lim_{x\to c}v(x)), we have (\lim_{x\to-\infty}2^{x}-\lim_{x\to-\infty}1) (\lim_{x\to-\infty}2^{x}=0) and (\lim_{x\to-\infty}1 = 1), so (\lim_{x\to-\infty}f(x)=0 - 1=-1)

Step3: Analyze the behavior as (x\to\infty)

For the exponential function (y = 2^{x}), when (x\to\infty), (2^{x}\to\infty) For (f(x)=2^{x}-1), (\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(2^{x}-1)) Using the limit rule (\lim_{x\to c}(u(x)-v(x))=\lim_{x\to c}u(x)-\lim_{x\to c}v(x)), we have (\lim_{x\to\infty}2^{x}-\lim_{x\to\infty}1) Since (\lim_{x\to\infty}2^{x}=\infty) and (\lim_{x\to\infty}1 = 1), then (\lim_{x\to\infty}(2^{x}-1)=\infty)

Answer:

a. ((0,0)) b. As (x\to-\infty), (f(x)\to - 1); As (x\to\infty), (f(x)\to\infty)