Question

for the function ( f(x)=2^{x}-1 ), find:
a. the ( y )-intercept.
( (0, quad) )
b. the behavior of the function as ( x ) approaches ( pm infty ).
as ( x
ightarrow-infty, f(x)
ightarrow )
as ( x
ightarrow infty, f(x)
ightarrow )

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 \(a^{0}=1\) (\(a
eq0\)), 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 \(x\to\infty\), \(f(x)\to\infty\)