Question

exercise 4.2.4
given the graph of the transformed exponential function, find a formula and describe the long run behavior.

f(x) =
long run behavior:
as x → -∞, f(x) →
as x → ∞, select an answer

Explanation:

Step1: Recall the general form of exponential function

The general form of an exponential function is \(y = ab^{x - h}+k\). The horizontal asymptote is \(y = k\). From the graph, the horizontal asymptote is \(y = 0\) (as \(x\to-\infty\), the function approaches \(0\)), so \(k = 0\). When \(x = 0\), \(y=2\). Let's assume the function is of the form \(y = ab^{x}\).

Step2: Substitute \(x = 0\) into the function

When \(x = 0\), \(y=ab^{0}=a\). Since \(y(0)=2\), then \(a = 2\).

Step3: Find another point to check the base \(b\)

Let's assume another point. Suppose we consider the point \((- 1,y)\). From the graph, when \(x=-1\), \(y = 1\). Substitute into \(y = 2b^{x}\), we get \(1=2b^{-1}\), then \(b = 2\). So the function is \(f(x)=2\cdot2^{x}=2^{x + 1}\)

Step4: Analyze the long - run behavior

For the exponential function \(y = 2^{x+1}\), as \(x\to-\infty\), \(2^{x+1}=\frac{2^{x}\cdot2}{1}\). Since \(\lim_{x\to-\infty}2^{x}=0\), then \(\lim_{x\to-\infty}f(x)=0\). As \(x\to\infty\), \(2^{x + 1}\to\infty\)

Answer:

\(f(x)=2^{x + 1}\)

Long run behavior:

As \(x\to-\infty\), \(f(x)\to0\)

As \(x\to\infty\), \(f(x)\to\infty\)