Question

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 an exponential function

The general form of an exponential function is \(y = ab^{x + h}+k\). When \(h = 0\) and \(k = 0\), it is \(y=ab^{x}\). We know that the \(y -\)intercept of the function \(y = ab^{x}\) is at \(x = 0\), and \(y=a\). From the graph, when \(x = 0\), \(y = 2\), so \(a = 2\).
We also know that when \(y = 0\), \(x=-2\). Substitute \(y = 0\), \(x=-2\) and \(a = 2\) into \(y=ab^{x}\): \(0 = 2b^{- 2}\) (This is wrong, we should use another approach. Let's assume the function is \(y = 2\cdot2^{x}\) (since when \(x = 0\), \(y=2\) and if we check another point. Let's use the fact that the function passes through \((-2,0)\) is wrong. Wait, no, the horizontal asymptote is \(y = 0\). Let's use two - point form. Let the function be \(y = ab^{x}\). We know \((0,2)\) gives \(a = 2\). Let's assume another point. When \(x=-1\), assume \(y = 1\) (from the graph's trend). Substitute \(x=-1\), \(y = 1\) and \(a = 2\) into \(y=ab^{x}\), we get \(1=2b^{-1}\), then \(b = 2\).

Step2: Write the formula of the function

The formula of the exponential function is \(f(x)=2\cdot2^{x}=2^{x + 1}\)

Step3: Analyze the long - run behavior

For an exponential function \(y = ab^{x}\) with \(a>0\) and \(b>1\) (here \(a = 2\), \(b = 2\)):
As \(x\to-\infty\), \(b^{x}=\frac{1}{b^{-x}}\to0\) (since \(b>1\) and \(-x\to+\infty\) as \(x\to-\infty\)). So \(f(x)=2\cdot2^{x}\to0\) as \(x\to-\infty\)
As \(x\to+\infty\), \(2^{x}\to+\infty\), so \(f(x)=2\cdot2^{x}\to+\infty\)

Answer:

\(f(x)=2^{x + 1}\); As \(x\to-\infty\), \(f(x)\to0\); As \(x\to\infty\), \(f(x)\to\infty\)