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Learning Objectives
 Graph exponential functions and theirtransformations.
 Review Laws of Exponents
Exponential functions are used for many realworld applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a realworld situation gives us a method for making predictions. Seeing their graphs gives us another layer of insight for predicting future events.
Graph BasicExponential Functions
Exponential growth is modelled by functions of the form \(f(x)=b^x\) where the base is greater than one. Exponential decay occurs when the base is between zero and one. We’ll use the functions \(f(x)=2^x\) and \(g(x)={\left(\tfrac{1}{2}\right)}^x\) to get some insight into the behaviour of graphs that model exponential growth and decay. In each table of values below, observe how the output values change as the input increases by \(1\).
Exponential Growth:\(f(x)=2^x\).
Each output value is the product of the previous output and the base, \(2\). We call the base \(2\) the constant ratio. In fact, for any exponential function with the form \(f(x)=ab^x,\) \(b\) is the constant ratio of the function. This means that as the input increases by \(1\), the output value \(f(x+1)\) will be the product of the base and the previous output, \(b f(x)=b \cdotab^x=ab^{x+1}=f(x+1)\), regardless of the value of \(a\). Notice from the table that
The domain of \(f(x)=2^x\) is all real numbers, the range is \((0,\infty)\), and the horizontal asymptote is \(y=0\).  Figure \(\PageIndex{1}\): Graph of the exponential growth function \(f(x)=2^x\). 
Exponential Decay:\(g(x)={\left(\frac{1}{2}\right)}^x\).
Again, as the input is increases by \(1\), each output value is the product of the previous output and the base, \(\tfrac{1}{2}\), so the constant ratio of the function is \(\tfrac{1}{2}\). Notice from the table that
The domain of \(g\) is all real numbers, the range is \((0,\infty)\), and the horizontal asymptote is \(y=0\).  Figure \(\PageIndex{2}\):Graph of the exponentialdecay function, \(g(x)={\left(\frac{1}{2}\right)}^x\). 
Notice that\(g(x)={\left(\frac{1}{2}\right)}^x = (2^{1})^x = f(x)\). Thusthe graph of \(g\) is simply a reflection over the \(y\)axis of the graph of \(f\).
CHARACTERISTICS OF THE GRAPH OF THE PARENT FUNCTION \(f(x) = b^x\)
An exponential function with the form \(f(x)=b^x\), \(b>0\), \(b≠1\), has these characteristics:
 Figure \(\PageIndex{3}\): Graphs of exponential growth and decay functions. 
How to: Graph a basicexponential function of the form \( y=b^x\)
 Draw and label the horizontal asymptote, \( y=0. \)
 Create a table of points and use it to plot at least \(3\) points, including the \(y\)intercept \( (0, 1) \) and key point \( (1, b) \).
 Draw a smooth curve that goes through the points and approaches the horizontal asymptote.
 State the domain \((−\infty,\infty)\), the range \((0,\infty)\), and the horizontal asymptote, \( y=0. \)
Example \(\PageIndex{1}\): Sketchthe Graph of an Exponential Function of the Form \(f(x) = b^x\)
Sketch a graph of \(f(x)=0.25^x\). State the domain, range, and asymptote.
Solution
Before graphing, identify the behavior and create a table of points for the graph.
\(x\)  \(−3\)  \(−2\)  \(−1\)  \(0\)  \(1\)  \(2\)  \(3\) 

\(f(x)={0.25}^x\)  \(64\)  \(16\)  \(4\)  \(1\)  \(0.25\)  \(0.0625\0  \(0.015625\) 
 Since \(b=0.25\) is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote \(y=0\).
 Create a table of points as in Table \(\PageIndex{3}\).
 Plot the asymptote, and the yintercept\((0,1)\), along with two other points. We can use \((−1,4)\) and \((1,0.25)\).
Draw a smooth curve connecting the points as shown to the right.
The domain is \((−\infty,\infty)\); the range is \((0,\infty)\); the horizontal asymptote is \(y=0\).
Try It \(\PageIndex{1}\)
Sketch the graph of \(f(x)=4^x\). State the domain, range, and asymptote 

Graph Transformations of Exponential Functions
Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function \(f(x)=b^x\) without loss of general shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied.
Vertical Shifts
The first transformation occurs when we add a constant \(d\) to the parent function \(f(x)=b^x\), giving us a vertical shift \(d\) units in the same direction as the sign. For example, if we begin by graphing a parent function, \(f(x)=2^x\), we can then graph two vertical shifts alongside it, using \(d=3\): the upward shift, \(g(x)=2^x+3\) and the downward shift, \(h(x)=2^x−3\). Both shifts areshown in the figure to the right.
Observe the results of shifting \(f(x)=2^x\) vertically:
 The domain, \((−\infty,\infty)\) remains unchanged.
 When the function is shifted up \(3\) units to \(g(x)=2^x+3\):
 The \(y\)intercept shifts up \(3\) units to \((0,4)\).
 The asymptote shifts up \(3\) units to \(y=3\).
 The range becomes \((3,\infty)\).
 When the function is shifted down \(3\) units to \(h(x)=2^x−3\):
 The \(y\)intercept shifts down \(3\) units to \((0,−2)\).
 The asymptote also shifts down \(3\) units to \(y=−3\).
 The range becomes \((−3,\infty)\).
Horizontal Shifts
The next transformation occurs when we add a constant \(c\) to the input of the parent function \(f(x)=b^x\), giving us a horizontal shift \(c\) units in the opposite direction of the sign. For example, if we begin by graphing the parent function \(f(x)=2^x\), we can then graph two horizontal shifts alongside it, using \(c=3\): the shift left, \(g(x)=2^{x+3}\), and the shift right, \(h(x)=2^{x−3}\). Both horizontal shifts are shown in the figure to the right.
Observe the results of shifting \(f(x)=2^x\) horizontally:
 The domain, \((−\infty,\infty)\), remains unchanged.
 The asymptote, \(y=0\), remains unchanged.
 The \(y\)intercept shifts such that:
 When the function is shifted left \(3\) units to \(g(x)=2^{x+3}\), the \(y\)intercept becomes \((0,8)\). This is because \(2^{x+3}=(8)2^x\), so the initial value of the function is \(8\).
 When the function is shifted right \(3\) units to \(h(x)=2^{x−3}\), the \(y\)intercept becomes \((0,\frac{1}{8})\). Again, see that \(2^{x−3}=(\frac{1}{8})2^x\), so the initial value of the function is \(\frac{1}{8}\).
SHIFTS OF THE PARENT FUNCTION \( y= b^x\)
For any constants \(c\) and \(d\), the function \(f(x)=b^{x+c}+d\) shifts the parent function \(y=b^x\)
 Vertically \(d\) units, in the same direction of the sign of \(d\).
 If \(d>0\) the parent function is shifted up \(d\) units
 If \(d<0\) the parent function is shifted down \(d\) units
 The new \(y\)coordinates are equal to \(y+d\)
 The horizontal asymptote becomes \(y=d\), and occurs when the exponential part of the function,\(b^{x+c}\) approaches zero.
 The range becomes \((d,\infty)\).
 Horizontally \(c\) units, in the opposite direction of the sign of \(c\).
 If \(c>0\) the parent function is shifted left\(c\) units
 If \(c<0\) the parent function is shifted right\(c\) units
 The new \(x\)coordinates are equal to \(xc\).;
 The newyintercept, at\((0,1)\) in the parent function,occuring at \(f(0)\), becomes \((0,b^c+d)\).
 The domain, \((−\infty,\infty)\), remains unchanged.
How to: Graph an exponential function of the form \( f(x)=b^{x+c}+d\)
 Draw the horizontal asymptote \(y=d\).
 Identify the shift as \((−c,d)\). Shift the graph of \(f(x)=b^x\) left \(c\) units if \(c\) is positive, and right \(c\) units if \(c\) is negative.
 Shift the graph of \(f(x)=b^x\) up \(d\) units if \(d\) is positive, and down \(d\) units if \(d\) is negative.
 State the domain, \((−\infty,\infty)\), the range, \((d,\infty)\), and the horizontal asymptote \(y=d\).
Example \(\PageIndex{2}\): Graphing a Shift of an Exponential Function
Graph \(f(x)=2^{x+1}−3\). State the domain, range, and asymptote. Solution We have an exponential equation of the form \(f(x)=b^{x+c}+d\), with \(b=2\), \(c=1\), and \(d=−3\). Draw the horizontal asymptote \(y=d\), so draw \(y=−3\). Identify the shift as \((−c,d)\), so the shift is \((−1,−3)\). Shift the graph of \(f(x)=b^x\) left \(1\) units and down \(3\) units. The domain is \((−\infty,\infty)\); the range is \((−3,\infty)\); the horizontal asymptote is \(y=−3\). 
Try It \(\PageIndex{2}\)
Graph \(f(x)=2^{x−1}+3\). State domain, range, and asymptote. 

Reflections
In addition to shifting, compressing, and stretching a graph, we can also reflect it about the \(x\)axis or the \(y\)axis. When we multiply the parent function \(f(x)=b^x\) by \(−1\), we get a reflection about the \(x\)axis. When we multiply the input by \(−1\), we get a reflection about the \(y\)axis. For example, if we begin by graphing the parent function \(f(x)=2^x\), we can then graph the two reflections alongside it. The reflection about the \(x\)axis, \(g(x)=−2^x\), is illustrated below in the graph on the left, and the reflection about the \(y\)axis \(h(x)=2^{−x}\), is shown in the graphon the right.
REFLECTIONS OF THE PARENT FUNCTION \(f(x) = b^x\)
The function \(f(x)=−b^x\)
 reflects the parent function \(f(x)=b^x\) about the \(x\)axis.
 has a \(y\)intercept of \((0,−1)\).
 has a range of \((−\infty,0)\)
 has a horizontal asymptote at \(y=0\) and domain of \((−\infty,\infty)\), which are unchanged from the parent function.
The function\(f(x)=−b^x + d\) has both a vertical shift and reflection about the \(x\)axis. In this situation, always do the vertical shift LAST.
The function \(f(x)=b^{−x}\)
 reflects the parent function \(f(x)=b^x\) about the \(y\)axis.
 has a \(y\)intercept of \((0,1)\), a horizontal asymptote at \(y=0\), a range of \((0,\infty)\), and a domain of \((−\infty,\infty)\), which are unchanged from the parent function.
The function\(f(x)=b^{x+c}\) has both a horizontalshift and reflection about the \(y\)axis. In this situation, always do the horizontal shift FIRST.
Example \(\PageIndex{3}\): Construct an Equation for a Reflected Exponential Function
Find and graph the equation for a function, \(g(x)\), that reflects \(f(x)=( \tfrac{1}{4} )^x\) about the \(x\)axis. State its domain, range, and asymptote.
Solution Since we want to reflect the parent function \(f(x)=( \tfrac{1}{4} )^x\) about the \(x\)axis, we multiply \(f(x)\) by \(−1\) to get, \(g(x)=−f(x)=−( \tfrac{1}{4} )^x\). Next we create a table of points below.
Plot the \(y\)intercept, \((0,−1)\), along with two other points. We can use \((−1,−4)\) and \((1,−0.25)\).Draw a smooth curve connecting the points.The domain is \((−\infty,\infty)\); the range is \((−\infty,0)\); the horizontal asymptote is \(y=0\).  Figure \(\PageIndex{11}\). Graph of \(g(x)=−( \tfrac{1}{4} )^x\). 
Try It \(\PageIndex{3}\)
Find and graph the equation for a function, \(g(x)\), that reflects \(f(x)={1.25}^x\) about the \(y\)axis. State its domain, range, and asymptote.
 Answer

\(g(x) = f(x) = 1.25^{x}\)
The domain is \((−\infty,\infty)\); the range is \((0,\infty)\); the horizontal asymptote is \(y=0\).
Included in the graph is the horizontal asymptote \(y=0\), and the points for \(g(1) = 1.25\), \(g(0)=1\), and \(g(1) = .8\).
Vertical Stretches or Compressions
While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression occurs when we multiply the parent function \(f(x)=b^x\) by a constant \(a>0\).
For example, if we begin by graphing the parent function \(f(x)=2^x\), we can then graph the stretch, using \(a=3\), to get \(g(x)=3{(2)}^x\) as shown in Figure (a), and the compression, using \(a=\dfrac{1}{3}\), to get \(h(x)=\dfrac{1}{3}{(2)}^x\) as shown on the right in Figure (b).
VERTICAL STRETCHES AND COMPRESSIONS OF THE PARENT FUNCTION \(y= b^x\)
For any factor \(a \ne 0\), the function \(f(x)=a{(b)}^x\)
 is stretched vertically by a factor of \(a\) if \(a>1\).
 is compressed vertically by a factor of \(a\) if \(0 < a<1\).
 The new \(y\)coordinates are equal to \(ay\). This would include vertical reflection if present.
 has a \(y\)intercept of \((0,a)\).
 has a horizontal asymptote at \(y=0\), a range of \((0,\infty)\), and a domain of \((−\infty,\infty)\), which are unchanged from the parent function.
If a vertically stretched, compressed and/or reflected function also has a vertical shift, like \(g(x)=a{(b)}^x + d, \) then the vertical shift, (\(d\) units up or down), must be done AFTER performing the vertical stretching, compression, and/or reflection.
Example \(\PageIndex{4}\): Graphing the Vertical Stretch of an Exponential Function
Sketch a graph of \(f(x)=4{\Big(\dfrac{1}{2}\Big)}^x\). State the domain, range, and asymptote.
Solution
Before graphing, identify the behavior and key points on the graph.
 Since \(b=\dfrac{1}{2}\) is between zero and one, the left tail of the graph will increase without bound as \(x\) decreases, and the right tail will approach the \(x\) axis as \(x\) increases.
 Since \(a=4\), the graph of the parent function \(y={\Big(\dfrac{1}{2}\Big)}^x\) will be stretched vertically by a factor of \(4\).
 Create a table of points of the parent function as shown in Table \(\PageIndex{4}\). Then multiply all the\(y\) coordinates by 4.
Table \(\PageIndex{4}\) \(x\) \(−3\) \(−2\) \(−1\) \(0\) \(1\) \(2\) \(3\) \( y={\Big(\dfrac{1}{2}\Big)}^x\) \(8\) \(4\) \(2\) \(1\) \(\dfrac{1}{2}\) \(\dfrac{1}{4}\) \(\dfrac{1}{8}\) \(f(x)=4{\Big(\dfrac{1}{2}\Big)}^x\)
\(32\) \(16\) \(8\) \(4\) \(2\) \(1\) \(0.5\)
 Plot the \(y\)intercept, \((0,4)\), along with two other points. We can use \((−1,8)\) and \((1,2)\).
 Draw a smooth curve connecting the points, as shown in the figure on the right.
The domain is \((−\infty,\infty)\); the range is \((0,\infty)\); the horizontal asymptote is \(y=0\).
Try It \(\PageIndex{4}\)
Sketch the graph of \(f(x)=\dfrac{1}{2}{(4)}^x\). State the domain, range, and asymptote.
 Answer

The domain is \((−\infty,\infty)\); the range is \((0,\infty)\); the horizontal asymptote is \(y=0\).
Combining Transformationsof Exponential Functions
Now that we have worked with each type of translation for the exponential function, we can summarize them.
How to: GRAPH TRANSFORMATIONSOF EXPONENTIAL FUNCTIONS
A transformation of an exponential function has the form\(f(x)=ab^{mx+c}+d\)
The transformations to the parent function, \(y=b^x\), \(b>1\), needed to obtain \(f\) are given below. The order these are done is important.
 FIRST. Shift horizontally \(c\) units. \( \quad x \rightarrow(x  c)\).
 If \(c > 0\) shift left \(c\) units.
 If \(c < 0\) shift right \(c\) units..
 Horizontal reflection, compression and/or stretching. \( \quad \) \( x \rightarrow\frac{1}{m} x \).
 If \( m \) is negative,reflect over the \(y\)axis. \(\quad \) \( (x \rightarrowx ).\)
 If \( m > 1 \) shrink the graph horizontally by a factor of \( \frac{1}{m}. \)
 If \( 0<m<1 \) stretchthe graph horizontally by a factor of \( \frac{1}{m}. \)
 Vertical reflection, compression and/or stretching. \(\quad \) \( y \rightarroway \).
 If \(a\) is negative, reflect over the \(x\)axis.\(\quad \) \( ( y \rightarrowy ) \).
 If \(a>0\) stretch the graph vertically by a factor of \( a\).
 If \(0<a<1\)compress the graph vertically by a factor of \( a\).
 LAST. Shiftvertically \(d\) units.\( \quad \) \( y \rightarrowy + d \).
 If\(d > 0\) shift up \(d\) units.
 If\(d < 0\)shift down \(d\) units.
If the exponential function is written inthe form \(f(x)=ab^{m(x + c)}+d\)then reverse the order ofsteps 1 and 2  do reflecting and stretching first, then do the horizontal shift.
Example \(\PageIndex{5}\)
Sketch the graphs of\( g(x) = 2 \cdot 3^{x5}+6 \) and \( f(x) = 3 \cdot 2^{x+1}4 \) and the corresponding original parent function. State the domain, range, and horizontal asymptote of the transformation.
Solution
The basic parent function for \( g(x) = 2 \cdot 3^{x5}+6 \)is \( y = 3^x\).
The transformations needed to obtain the graph of \(g(x)\) from the graph of \(y\) are:
 Shift right 5units ( \( x \rightarrowx+5\) )
 Reflect over the \(x\)axis ( \( y \rightarrowy\) )
 Vertically stretch by a factor of 2 ( \( y \rightarrow2y\) )
 Shift up \(6\) units ( \( y \rightarrowy4\) )
The graph of \(g(x)\) and its parent function is on the right.
The domain is \((−\infty,\infty)\); the range is \((\infty,6)\); the horizontal asymptote is \(y=6\).
If tables are used to graph the function, coordinate points for the parent function appear in the table below
\( \begin{array}{rcccc c }
\hline\text{Parent function: }x& \text{HA}& 1& 0 &1& 2\\
\hline y= 3^x & 0 & \frac{1}{3}& 1 & 3 & 9 \\[2pt]
\hline
\end{array} \)
Corresponding coordinate points for the transformation \( g(x) = 2 \cdot 3^{x5}+6 \) would be
\( \begin{array}{rcccc c }
\hline\text{New }x \text{ is: } \:\;\quad x+5& \text{HA}& 4& 5 &6& 7\\
\hline\text{New } y \text{ is: } 2y+6 & 6 & 5\frac{1}{3}& 4 & 0 & 12 \\[2pt]
\hline
\end{array} \)
The basic parent function for \( f(x) = 3 \cdot 2^{x+1}4 \) is \( y = 2^x\).
The transformations needed to obtain the graph of \(f(x)\) from the graph of \(y\) are:
 Shift left 1 unit ( \( x \rightarrowx1 \) )
 Reflect over the \(y\)axis ( \( x \rightarrowx\) )
 Vertically stretch by a factor of 3 ( \( y \rightarrow3y\) )
 Shift down \(4\) units ( \( y \rightarrowy4\) )
If instead we rewrote the function in the form: \( f(x) = 3 \cdot 2^{(x1)}4 \), the transformations would be
 Reflect over the \(y\)axis ( \( x \rightarrowx\) )
 Shift right1 unit ( \( x \rightarrowx+1 \) )
 Vertically stretch by a factor of 3 ( \( y \rightarrow3y\) )
 Shift down \(4\) units ( \( y \rightarrowy4\) )
The graph of \(f(x)\) and its parent function is on the right.
The domain is \((−\infty,\infty)\); the range is \((\infty, 4)\); the horizontal asymptote is \(y=4\).
If tables are used to graph the function, coordinate points for the parent function appear in the table below
\( \begin{array}{rcccc c }
\hline\text{Parent function: }x& \text{HA}& 1& 0 &1& 2\\
\hline y= 2^x & 0 & \frac{1}{2}& 1 & 2 & 4 \\[2pt]
\hline
\end{array} \)
Corresponding coordinate points for the transformation \( f(x) = 3 \cdot 2^{x+1}4 \) would be
\( \begin{array}{rcccc c }
\hline\text{New } x \text{ is: } x+1& \text{HA}& 2& 1 &0& 1\\
\hline\text{New } y \text{ is: } 3y4& 4 & 2\frac{1}{2}& 1 & 2 & 8\\[2pt]
\hline
\end{array} \)
Example \(\PageIndex{6}\)
Sketch the graph of \( f(x) = 2 ^ {\frac{1}{2} x + 4}  3 \).State the domain, range, and horizontal asymptote of the transformation.
Solution
The basic parent functionis \( y = 2^x\).
The transformations needed to obtain the graph of \(f(x)\) from the graph of \(y\) are:
 Shiftleft \(4\)units ( \( x \rightarrowx4\) )
 Horizontally stretch by a factor of 2 ( \( x \rightarrow2x\) )
 Shift down \(3\) units ( \( y \rightarrowy3\) )
The graph of \(f(x)\) and its parent function is on the right.
The domain is \((−\infty,\infty)\); the range is \((3, \infty)\); the horizontal asymptote is \(y=3\).
If tables are used to graph the function, coordinate points for the parent function appear in the table below
\( \begin{array}{rcccc c }
\hline\text{Parent function: }x& \text{HA}& 1& 0 &1& 2\\
\hline y= 2^x & 0 & \frac{1}{2}& 1 & 2 & 4 \\[2pt]
\hline
\end{array} \)
Corresponding coordinate points for the transformation would be
\( \begin{array}{rcccc c }
\hline\text{New }x \text{ is: } 2(x4)& \text{HA}& 10& 8 &6& 4\\
\hline\text{New } y \text{ is: } y3& 3 & 2.5& 2 & 1 & 1\\[2pt]
\hline
\end{array} \)
Note that if the function were rewritten in the form:\( f(x) = 2 ^ {\frac{1}{2} (x + 8)}  3 \) then the transformations indicated by this version of the function would be
 Horizontally stretch by a factor of 2 ( \( x \rightarrow2x\) )
 Shiftleft \(8\)units ( \( x \rightarrowx8\) )
 Shift down \(3\) units ( \( y \rightarrowy3\) )
Try It \(\PageIndex{7}\)
Sketch the graph of \( f(x) = 2 ^ {2x  4} + 3 \).State the domain, range, and horizontal asymptote of the transformation.
 Answer

The domain is \((−\infty,\infty)\); the range is \((3, \infty)\); the horizontal asymptote is \(y=3\).
As written the transformations done on the parent function \(y = 2^x\) would be
\( \quad \) right 4, reflectover y axis, \( x \rightarrow\frac{1}{2} x \), up 3.
However, if rewritten as \( f(x) = 2 ^ {2 (x +2)} + 3 \), the transformations would be
\( \quad \) reflectover y axis, \( x \rightarrow\frac{1}{2} x \), left 2, up 3.
Construct an Exponential Equation from a Description
Example \(\PageIndex{8}\): Writea Function from a Description
Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range.
\(f(x)=e^x\) is vertically stretched by a factor of \(2\) , reflected across the yaxis, and then shifted up \(4\) units.
Solution
We want to find an equation of the general form \(f(x)=ab^{x+c}+d\). We use the description provided to find \(a\), \(b\), \(c\), and \(d\).
 We are given the parent function \(f(x)=e^x\), so \(b=e\).
 The function is vertically stretched by a factor of \(2\), so \(a=2\).
 The function is reflected about the \(y\)axis. We replace \(x\) with \(−x\) to get: \(e^{−x}\).
 The graph is shifted vertically \(4\) units, so \(d=4\).
Substituting in the general form we get,
\(f(x)=ab^{x+c}+d\)
\(=2e^{−x+0}+4\)
\(=2e^{−x}+4\)
The domain is \((−\infty,\infty)\); the range is \((4,\infty)\); the horizontal asymptote is \(y=4\).
Try It \(\PageIndex{8}\)
Write the equation for function described below. Give the horizontal asymptote, the domain, and the range.
\(f(x)=e^x\) is compressed vertically by a factor of \(\dfrac{1}{3}\), reflected across the \(x\)axis and then shifted down \(2\) units.
 Answer

\(f(x)=−\dfrac{1}{3}e^{x}−2\); the domain is \((−\infty,\infty)\); the range is \((−\infty,2)\); the horizontal asymptote is \(y=2\).
Exponent Properties
When learning about exponents, we are typically given expressions that have a whole number exponent and a base that is a variable, like \(x^2\). These type of expressions are technically called power functions.In contrast, the base can be a constant and the exponent can have a variable in it, like \(2^x\). Expressions inthis form are called exponential functions. When simplifying expressions with the variable in the exponent, we often use the Laws of Exponents "backwards". A review of properties of exponentsfrom a different perspective is briefly discussed here. Competency with exponents is a very useful skill to have. The table below summarizes these properties.
Rules of Exponents
For nonzero real numbers aand band integers m and n
Product Rule  Quotient Rule  Power Rule  

Like Bases Rule  \(a^m⋅a^n=a^{m+n}\)  \( \dfrac{a^m}{a^n}=a^{m−n}\)  \((a^m)^n=a^{m⋅n}\) 
Reverse of Like Bases Rule  \(a^{m+n}=a^m⋅a^n\)  \(a^{m−n}= \dfrac{a^m}{a^n}\)  \(a^{m⋅n}=(a^m)^n\) 
Like Exponents Rule  \(a^n⋅b^n =(a⋅b)^n \)  \(\dfrac{a^n}{b^n} = \left(\dfrac{a}{b}\right)^n\)  \(\dfrac{1}{b^n} = \left(\dfrac{1}{b}\right)^n\) 
Example \(\PageIndex{9}\)
Rewrite the following exponential expressions such that the result uses only one exponent, \(x\). Simplify.
1. \(2^{x+3}\)  2. \(3^{x2}\)  3. \(5^{3x}\)  4. \(2^x\cdot11^x\)  5. \(5^{3+2x} \)  6. \( \dfrac{4^x}{12^x} \)  7. \( \dfrac{6^{3x}}{9^{2x}} \)  8. \( \dfrac{8^{x1}}{2^{2x5}} \) 
Solution
1. \(2^{x+3} = 2^x 2^3 = 2^x \cdot8 = 8(2^x)\) 2. \(3^{x2} = 3^x3^{2} = \dfrac{3^x}{3^2} = \dfrac{3^x}{9} \) 3. \(5^{3x} = 5^3 5^{x} = \dfrac{5^3}{5^x} = \dfrac{125}{5^x} = 125 \left( \dfrac{1}{5} \right)^x \) 4. \(2^x\cdot11^x = (2 \cdot11)^x = 22^x \) 5. \(5^{3+2x} = 5^3 5^{2x} = 125 (5^2)^x = 125(25^x) \)  6. \( \dfrac{4^x}{12^x} =\left( \dfrac{4}{12} \right)^x =\left( \dfrac{1}{3} \right)^x \) 7. \( \dfrac{6^{3x}}{9^{2x}} = \dfrac{(6^3)^x}{(9^2)^x} =\left( \dfrac{6^3}{9^2} \right)^x =\left( \dfrac{8}{3} \right)^x \) 8. \( \dfrac{8^{x1}}{2^{2x5}} =\dfrac{8^x 8^{1}}{2^{2x} 2^{5}} = \dfrac{8^x}{2^{2x}} \cdot \dfrac{2^5}{8}=\dfrac{2^5}{8} \cdot \dfrac{8^x}{4^x}= \dfrac{32}{8} \cdot \left( \dfrac{8}{4}\right)^x \\ \) 
\(\PageIndex{9}\)
Rewrite the following exponential expressions such that the result uses only one exponent, \(x\). Simplify.
\(1.\quad 3^{x+1}\cdot4^{2x} \)  \(2. \quad \dfrac{2^{3x+4}}{7^{2x+1}} \) 
 Answer

\( 1.\quad 3 \left( \dfrac{3}{16} \right) ^x \qquad2. \quad\dfrac{16}{7}\left( \dfrac{8}{49} \right) ^x \)
Try It \(\PageIndex{9b}\)
Use exponent properties to describe the transformations needed to graph the function \(f(x) = 5 \cdot 3^{4x+2}6\)from a parent function.
 Answer
 \(f(x) = 5 \cdot 3^{4x+2}6\) =\(f(x) = 5 \cdot 3^{4x} \cdot3^{2}6\)=\(f(x) = 5 \cdot3^{2}\cdot(3^4)^{x} 6\) =\(f(x) =45\cdot81^{x} 6\)
 Parent function is now \(y = 81^x\)
Reflect over the \(y\)axis ( \( x \rightarrowx\) )
Vertically stretch by a factor of 45 ( \( y \rightarrow45y\) )
Shift down \(6\) units ( \( y \rightarrowy6\) )
Key Concepts
General Form for the Translation of the Parent Function \(f(x)=b^x\)  \(f(x)=ab^{x+c}+d\) or \(f(x)=ab^{x+c}+d\) 
 The graph of the function \(f(x)=b^x\) has a yintercept at \((0, 1)\),domain \((−\infty, \infty)\),range \((0, \infty)\), and horizontal asymptote \(y=0\).
 If \(b>1\),the function is increasing. The left tail of the graph will approach the asymptote \(y=0\), and the right tail will increase without bound.
 If \(0<b<1\), the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote \(y=0\).
 The equation \(f(x)=b^x+d\) represents a vertical shift of the parent function \(f(x)=b^x\).
 The equation \(f(x)=b^{x+c}\) represents a horizontal shift of the parent function \(f(x)=b^x\).
 The equation \(f(x)=ab^x\), where \(a>0\), represents a vertical stretch if \(a>1\) or compression if \(0<a<1\) of the parent function \(f(x)=b^x\).
 When the parent function \(f(x)=b^x\) is multiplied by \(−1\), the result, \(f(x)=−b^x\), is a reflection about the xaxis. When the input is multiplied by \(−1\), the result, \(f(x)=b^{−x}\), is a reflection about the yaxis.
 All translations of the exponential function can be summarized by the general equation \(f(x)=ab^{mx+c}+d\).
 Using the general equation \(f(x)=ab^{x+c}+d\), we can write the equation of a function given its description.