Analyze \(f(x) = \frac {1}{3} \cdot \, 2^{x+5} - 7\)

Categorize: \(f(x) = \frac {1}{3} \cdot \, 2^{x+5} - 7\) is a shifted exponential function, since it matches our template, \(A \, r^{B \, x + C} + D\)

Let’s think about the pieces of the formula to help write our analysis.

Compared to a basic exponential function graph, where the horizontal axis is the horizontal asymptote, here, the horizontal asymptote has been moved down \(\answer {7}\) units, because the constant term in the formula is \(-7\).

The “inside” (the argument) of the formula for \(f(x)\), representing the domain, is \(x+5\). This equals \(0\), when \(x=-5\). The exponent is positive for \(x>-5\), and because the base is \(2 > 1\), this is the direction (right) of unbounded growth. Therefore, the other direction (left) is where the horizontal asymptote is in effect. Since the leading coefficient is \(\frac {1}{3} > 0\), the unbounded growth is positive.

At \(x=-5\), we have our one anchor point for the graph. The point is \(\left (-5, \frac {1}{3} - 7 \right )\), which is \(\frac {1}{3}\) above the horizontal asymptote, \(y = -7\).

Graph of \(y = f(x)\).

With these ideas, we can write an algebraic analysis.

Domain:

The domain of every shifted exponential function is \((-\infty , \infty )\).

Zeros:

\(f(x) = \frac {1}{3} \cdot \, 2^{x+5} - 7 = 0\)

\begin{align*} \frac {1}{3} \cdot \, 2^{x+5} - 7 & = 0 \\ \frac {1}{3} \cdot \, 2^{x+5} & = 7 \\ 2^{x+5} & = 21 \\ x+5 & = \log _2{21} \\ x & = \log _2{21} - 5 \end{align*}

Note: \(\log _2{21} - 5 \approx -0.607\), which agrees with the graph.

Continuity:

Shifted exponential functions are continuous.

Behavior:

Like exponential functions, shifted exponential functions are always increasing or always decreasing.

The leading coefficient of \(f\) is \(\frac {1}{3}\), which is positive.

The base of \(f\) is \(2 > 1\).

The leading coefficient of the argument is \(1\), which is also positive.

Comparing to \(e^x\), a base of \(2\) does not change the behavior. A positive leading coefficient does not chnage the behavior. A positive leading coefficient for the linear exponent does not change the behavior. We have an increasing function.

End-Behavior:

The negative direction is the direction that makes the exponent negative. Since the base is \(2 > 1\), that makes this the direction that the function value levels off to the constant term, \(-7\).

The other direction is where the shifted exponential function is uunbounded. Since the leading coefficient of \(f\) is positive, \(f\) is unbounded positively.

Global Extrema:

Shifted exponential functions do not have global maximums or minimums.

Local Extrema:

Shifted exponentials function do not have local maximums or minimums.

Range: The end-behavior tells us that the range is \((-7, \infty )\).

\(f(x) = \frac {1}{3} \, 2^{x+5} - 7\) is the function in the last example. We could take advantage of exponential algebra to simplify the formula.

We now have a basic exponential function shifted down \(7\). The new base is still \(2\) and the new leading coefficient is \(\frac {32}{3}\).

By doing this, we change our strategic point to \(\left ( 0, \frac {32}{3} - 7 \right ) = \left ( 0, \frac {11}{3} \right )\).

Let \(G(t) = 4 \cdot 3^{2t-1} - 5\).

We could take advantage of exponential algebra to simplify the exponent.

We now have a basic exponential function shifted down \(\answer {5}\). The new base is \(\answer {9}\) and the new leading coefficient is \(\answer {\frac {4}{3}}\).

The leading coefficient is \(\frac {4}{3} > 0\) and the base is \(9 > 1\). Therefore, \(G(t)\) is an increasing function. The function values will be positive, once they overcome the drop of \(5\) to the horizontal asymptote.

\(G(t) = \frac {4}{3} \cdot 9^t - 5\) is a transformation of \(g(t) = 3^t\). Let’s compare \(G(t)\) back to \(3^t\).

Since the new base \(9\) is greater than the old base of \(3\), the function grows faster. The graph goes up to the right faster and steeper. This means the graph is compressed horizontally, which is the result of multiplying by \(2\) in the exponent.

End-Behavior:

Analyze \(B(t) = -2 \, \left ( \frac {2}{3} \right )^{3-t} + 4\)

Categorize: \(B(t) = -2 \, \left ( \frac {2}{3} \right )^{3-t} + 4\) is a shifted exponential function, becuase it matches our template, \(A \, r^{B \, x + C} + D\).

Formula dissection:

\(\blacktriangleright \) the base: \(\frac {2}{3} < 1\)
\(\blacktriangleright \) the linear exponent: \(3-t\) is positive for large negative \(t\).
\(\blacktriangleright \) the linear exponent: \(3-t\) is negative for large positive \(t\).
\(\blacktriangleright \) the leading coefficient is negative.

Together, these tell us that \(B(t)\) settles down for large negative \(t\) and that \(B(t)\) becomes unbounded when \(t\) is large and positive.

This is a transformed version of the basic exponential function template \(\left ( \frac {2}{3} \right )^{-t} = \left ( \answer {\frac {3}{2}} \right )^t\).

When \(t < 0\), then \(-t > 0\) and we get \(\left ( \frac {2}{3} \right )^{positive}\) and the basic exponential portion is becoming smaller, approaching \(0\).

\[ \lim \limits _{t \to -\infty } \left ( \frac {2}{3} \right )^{-t} = 0 \]

When \(t > 0\), then \(-t < 0\) and we get \(\left ( \frac {2}{3} \right )^{negative} = \left ( \frac {3}{2} \right )^{positive}\) and \(B(t)\) is becoming unbounded.

\[ \lim \limits _{t \to \infty } \left ( \frac {2}{3} \right )^{-t} = \lim \limits _{t \to \infty } \left ( \frac {3}{2} \right )^t = \infty \]

Exponential growth to the right and decay to the left.

Since \(a = -2 < 0\), these smaller/larger values of \(-2 \, \left ( \frac {2}{3} \right )^{-t}\) are smaller/larger negative values.

We also have two shifts. One from the \(3\) in the exponent and the second from the constant term, \(4\):

\(\blacktriangleright \) Vertical Shift

Adding \(4\) to the outside shifts the graph vertically up \(4\). The asymptote is \(y = 4\) and

\[ \lim \limits _{t \to -\infty } B(t) = 4 \]

\(\blacktriangleright \) Horizontal Shift

Our exponent here is \(3 - t\). Our function’s exponent is zero, \(3-t=0\) when \(t=3\). Our one anchor point is shifted over to \(3\). Multipying by \(-2\), means the dot is \(2\) away(below) from the horizontal asymptote, which is now \(y=4\). Our anchor point is \((3, 2)\).

Graph of \(y = B(t)\).

With these ideas, we can write an algebraic analysis.

Domain:

The domain of every shifted exponential function is \((-\infty , \infty )\).

Zeros:

\(B(t) = -2 \, \left ( \frac {2}{3} \right )^{3-t} + 4\)

\begin{align*} -2 \, \left ( \frac {2}{3} \right )^{3-t} + 4 & = 0 \\ -2 \, \left ( \frac {2}{3} \right )^{3-t} & = -4 \\ \left ( \frac {2}{3} \right )^{3-t} & = \frac {-4}{-2} = 2 \\ 3-t & = \log _{\tfrac {2}{3}}(2)\\ 3 - \log _{\tfrac {2}{3}}(2) & = t \end{align*}

Note: We’ll need to switch bases to use the calculatr=or to get an approximatio to comapre with the graph.

That agrees with the graph.

Continuity:

Shifted exponential functions are continuous.

Behavior:

Like exponential functions, shifted exponential functions are always increasing or always decreasing.

The leading coefficient of \(B\) is \(-2\), which is negative.

The base of \(B\) is \(\frac {2}{3} < 1\).

The leading coefficient of the linear exponent is \(-1\), which is negative.

Comparing to \(e^x\), a base of \(-2\) does change the behavior from increasing to decreasing. A negative leading coefficient changes the behavior back to increasing. A negative leading coefficient for the linear exponent does changes the behavior again to decreasing. We have an decreasing function.

End-Behavior:

The positive direction is the direction that makes the exponent negative. Since the base is \(\frac {2}{3} < 1\), that makes this the direction that the function value becomes unbounded. The negative leading coefficient means \(B\) is unbounded negatively.

The other direction is where the shifted exponential function approached the constant term

Global Extrema:

Shifted exponential functions do not have global maximums or minimums.

Local Extrema:

Shifted exponentials function do not have local maximums or minimums.

Range: The end-behavior tells us that the range is \((-\infty , 4)\).

Consider \(K(f) = 3^{5-f} - 5\)

Graph of \(y = K(f)\).

Quick ideas:

• The natural or implied domain of \(K\) is \(\mathbb {R}\).
• \(K\) is always decreasing.
• \(K\) has no maximums or minimums.
• \(\lim \limits _{f \to -\infty } K(f) = \infty \)
• \(\lim \limits _{f \to \infty } K(f) = -5\)