The exponent gets big and positive when \(f\) gets big and positivenegative.
shifted range
The formula template for the basic exponential function looks like
\[ a \, r^x \, \text { with } \, a, r \in \mathbb {R} \, | \, r > 0 \]
As we have seen before, the coefficient \(a\) controls vertical stretching or compression. The sign of \(a\) dictates the sign of our function values. \(r\) dictates a growing or decaying function.
Shifted exponential functions shift the range by adding a constant.
\[ a \, r^x + b \, \text { with } \, a, b, r \in \mathbb {R} \, | \, r > 0 \]
These no longer have a constant percent growth rate. However, their analysis is
exactly the same as for exponential functions with one big difference in our
conclusions. Shifted exponential functions may have zeros.
Shifted Exponential Function
![[Picture]](shiftedExponentialFunctions-e8a66b3bd36ef056fe292c32674071c0.svg)
Analyze \(f(x) = \frac {1}{3} \, 2^{x+5} - 7\)
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 \(y = f(x)\).
- The natural or implied domain of \(f\) is \(\mathbb {R}\), because \(f\) is a shifted exponential
function.
- \(f\) is always increasing, because the base is greater than \(1\), and the leading
coefficient of the function and the leading coefficient of the exponent have
the same sign.
- \(f\) has no maximums or minimums, because shifted exponential functions
don’t have global or local maximums or minimums.
-
end-behavior
- \(\lim \limits _{x \to -\infty } f(x) = -7\)
- \(\lim \limits _{x \to \infty } f(x) = \infty \)
- This end-behavior gives us a range of \((-7, \infty )\).
Shifted Exponential Function
![[Picture]](shiftedExponentialFunctions-29bc0d0b5d559f3d1679b025ebb1e481.svg)
Analyze \(B(t) = -2 \, \left ( \frac {2}{3} \right )^{3-t} + 4\)
First, observe that \(\frac {2}{3} < \answer {1}\).
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.
\[ \lim \limits _{t \to \infty } \left ( \frac {2}{3} \right )^{-t} = \infty \]
Next, we have a negative leading coefficient.
Since \(-2 < 0\), the values of \(-2 \, \left ( \frac {2}{3} \right )^{power}\) are always negative.
Finally, we also have two shifts:
\(\blacktriangleright \) 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 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 \(y = B(t)\).
- The natural or implied domain of \(B\) is \(\mathbb {R}\), because \(B\) is a shifted exponential
function.
- \(B\) is always decreasing, because the base is less than \(1\), and the leading
coefficient of the function and the leading coefficient of the exponent have
the same sign.
- \(B\) has no maximums or minimums, because shifted exponential functions
don’t have global or local maximums or minimums.
-
end-behavior
- \(\lim \limits _{t \to -\infty } B(t) = 4\)
- \(\lim \limits _{t \to \infty } B(t) = -\infty \)
- This end-behavior gives us a range of \((-\infty , 4)\).
The graph agrees with our analysis.
Shifted Exponential Function
![[Picture]](shiftedExponentialFunctions-f83dbf3821e69cd6cc46fbf1e6cc0442.svg)
Analyze \(K(f) = 3^{5-f} - 5\)
Graph of \(y = K(f)\).
- The natural or implied domain of \(K\) is \(\mathbb {R}\), because \(K\) is a shifted exponential
function.
- \(K\) is always decreasing, because the base is greater than \(1\), and the leading
coefficient of the function and the leading coefficient of the exponent have
opposite signs.
- \(K\) has no maximums or minimums, because shifted exponential functions
don’t have global or local maximums or minimums.
-
end-behavior
- \(\lim \limits _{f \to -\infty } K(f) = \infty \)
- \(\lim \limits _{f \to \infty } K(f) = -5\)
- This end-behavior gives us a range of \((-5, \infty )\).
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more examples can be found by following this link
More Examples of Percent Change