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

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

For the basic exponential function graph, the horizontal axis is the horizontal asymptote. Here, this has been moved down \(7\).

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

\(f\) is the sum of an exponential function, \(\frac {1}{3} \, 2^{x+5}\), and a constant function, \(-7\). The exponential part will approach \(0\) on one side of the domain. Therefore, \(f\) will approach \(-7\) on that side of the domain.

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 create an algebraic analysis.

Domain

\(f\) is a shifted exponential function, which tells us that its domain is \((-\infty , \infty )\).

Zeros

\(f\) is a shifted exponential function, therefore it might have a zero.

Note: \(\log _2(21) - 5 \approx -0.6076825772\), which agrees with the graph.

Continuity

\(f\) is a shifted exponential function, therefore it is continuous.

Behavior (Increasing and Decreasing)

• The base is \(2\), which is greater than \(1\).
• The leading coeffcient is \(\frac {1}{3} > 0\).
• The leading coeffcient of the linear exponent is \(1 > 0\).

That tells us that \(f\) is increasing.

End-Behavior

\(f\) is a shifted exponential function, therefore the end-behavior of one side is the constant term, \(-7\) and the other is unbounded. The leading coefficient is \(\frac {1}{3}\), which tells is that when \(f\) becomes unbounded, it will become unbounded positively.

\(f\) is increasing and is unbounded positively. That gives us

Local Maximum and Minimum

\(f\) is a shifted exponential function, therefore it has no local extrema.

Global Maximum and Minimum

\(f\) is a shifted exponential function, therefore it has no global extrema.

Range

• \(f\) is continuous
• \(f\) is increasing
• \(\lim \limits _{x \to -\infty } f(x) = -7\)
• \(\lim \limits _{x \to \infty } f(x) = \infty \)

The range of \((-7, \infty )\).

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, since it matches our template, \(A \, r^{B \, x + C} +D\)

First, observe that the base is \(\frac {2}{3} < \answer {1}\).

Our base is less than \(1\). Therefore, as its exponent gets large and positive, we multiply by more \(\frac {2}{3}\)’s and the overall values get smaller.

Except, the variable, \(t\), in the exponent is multiplied by \(-1\). Therefore, we need \(t\) to get large and negative in order for the exponent to get large and positve.

• \(\left ( \frac {2}{3} \right )^{3-t}\) decays when \(t\) becomes more negative.
• \(\left ( \frac {2}{3} \right )^{3-t}\) grows when \(t\) becomes more positive.

The exponential stem of \(B(t)\) is \(\left ( \frac {2}{3} \right )^{-t}\), which is a transformed version of the basic exponential function model \(M(t) = \left ( \frac {2}{3} \right )^{t}\).

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

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

\(\blacktriangleright \) Graphing

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

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

Our exponent is \(3 - t = -t + 3\). Our anchor point for graphing is associated with the exponent equalling \(0\).

\(3-t=0\) when \(t=3\). Our one anchor point is shifted over to \(3\). Multipying by \(-2\), means the dot is \(2\) away from the horizontal asymptote, which is now \(y=4\).

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

With these ideas, we can create an algebraic analysis.

Domain

\(B\) is a shifted exponential function, therefore its domain is \((-\infty , \infty )\).

Zeros

\(B\) is a shifted exponential function, therefore it might have a zero.

Note: \(3 - \log _{\tfrac {2}{3}}(2) \approx -3.584962501\), which agrees with the graph.

Continuity

\(B\) is a shifted exponential function, therefore it is continuous.

End-Behavior

\(B\) is a shifted exponential function, therefore on one side the end-behavior is the constant term \(4\) and unbounded on the other side.

Since the leading coefficient is \(-2 < 0\), we know \(B\) will be unbounded negatively. We just need to figure out which side.

• the base of \(B\) is \(\frac {2}{3} < 1\)
• the leading coefficient of \(B\) is \(-2\), which is negative and tells us that \(B\) will become unbounded negatively.
• the leading coefficient of the linear exponent is \(-1\), which is negative.

This makes \(B\) a decreasing function, which becomes unbounded negatively. That gives us

Behavior (Increasing and Decreasing)

• the base of \(B\) is \(\frac {2}{3} < 1\)
• the leading coefficient of \(B\) is \(-2\), which is negative.
• the leading coefficient of the linear exponent is \(-1\), which is negative.

This makes \(B\) a decreasing function.

Local Maximum and Minimum

\(B\) is a shifted exponential function, therefore it has no local extrema.

Global Maximum and Minimum

\(B\) is a shifted exponential function, therefore it has no global extrema.

Range

• \(B\) is continuous
• \(\lim \limits _{t \to -\infty } B(t) = 4\)
• \(\lim \limits _{t \to \infty } B(t) = -\infty \)

The range is \((-\infty , 4)\).

The graph agrees with our analysis.

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

The formula for \(K\) matches which template?

\(A \, x + B\) \(A \, x^2 + B \, x + C\) \(A \, |B \, x + C | + D\) \(A \, r^{B \, x + C}\) \(A \, r^{B \, x + C} + D\) \(A \, \ln (B \, x + C) + D\)

The exponent gets big and positive when \(f\) gets big and positivenegative.

The graph will grow to the rightleft.

.

The graph will approach the asymptote to the rightleft.

Our one anchor point moves to \(\left (\answer {5}, \answer {-4}\right )\).

The graph will become unbounded updown.

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

With these ideas, we can create an algebraic analysis.

Domain

The natural or implied domain of \(K\) is \(\mathbb {R}\), because \(K\) is a shifted exponential function.

Zeros

\(B\) is a shifted exponential function, so it might have a zero.

Note: \(5 - \log _3(5) \approx 3.535026479\), which agrees with the graph.

Continuity

\(K\) is continuous, because \(K\) is a shifted exponential function.

End-Behavior

• The base is \(3 > 1\)
• The leading coefficient is \(1\), which is positive
• The leading coefficient of the linear exponent is \(-1\), which is negative

The base is greater than \(1\) and the leading coeffcients are of opposite sign. That tells us that \(K\) is decreasing.

The positive leading coefficient tells us that \(K\) becomes unbounded positively. We just need to figure out which side. The end-behavior on the other side is the constant term, \(-5\).

\(K\) is a decreasing shifted exponential function that becomes unbounded positively.

Behavior (Increasing and Decreasing)

The base is greater than \(1\) and the leading coeffcients are of opposite sign. That tells us that \(K\) is decreasing.

Global Maximum and Minimum

Shifted exponential functions do not have global maximum or minimum values.

Local Maximum and Minimum

Shifted exponential functions do not have local maximum or minimum values.

Range

• \(K\) is continuous
• \(K\) is decreasing.
• \(\lim \limits _{f \to -\infty } K(f) = \infty \)
• \(\lim \limits _{f \to \infty } K(f) = 0\)

The range is \((0, \infty )\).

This all agrees with the graph.