4.2: Graphs of Exponential Functions (2024)

How to: Graph an exponential function of the form \( f(x)=b^{x+c}+d\)

1. Draw the horizontal asymptote \(y=d\).
2. 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.
3. Shift the graph of \(f(x)=b^x\) up \(d\) units if \(d\) is positive, and down \(d\) units if \(d\) is negative.
4. 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

Try It \(\PageIndex{2}\)

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.

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.

1. FIRST. Shift horizontally \(c\) units. \( \quad x \rightarrow(x - c)\).
   • If \(c > 0\) shift left \(c\) units.
   • If \(c < 0\) shift right \(c\) units..
2. Horizontal reflection, compression and/or stretching. \( \quad \) \( x \rightarrow\frac{1}{m} x \).
   • If \( m \) is negative,reflect over the \(y\)-axis. \(\quad \) \( (x \rightarrow-x ).\)
   • 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 \( \frac{1}{|m|}. \)
3. Vertical reflection, compression and/or stretching. \(\quad \) \( y \rightarroway \).
   • If \(a\) is negative, reflect over the \(x\)-axis.\(\quad \) \( ( y \rightarrow-y ) \).
   • If \(|a|>0\) stretch the graph vertically by a factor of \( |a|\). 
   • If \(0<|a|<1\)compress the graph vertically by a factor of \( |a|\). 
4. 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\( g(x) = -2 \cdot 3^{x-5}+6 \) 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 \( g(x) = -2 \cdot 3^{x-5}+6 \)is \( y = 3^x\).

The transformations needed to obtain the graph of \(g(x)\) from the graph of \(y\) are:

1. Shift right 5units ( \( x \rightarrowx+5\) )
2. Reflect over the \(x\)-axis ( \( y \rightarrow-y\) )
3. Vertically stretch by a factor of 2 ( \( y \rightarrow2y\) )
4. Shift up \(6\) units ( \( y \rightarrowy-4\) )

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

\( \begin{array}{|r|c|c|c|c| c |}
\hline\text{Parent function: }x& \text{HA}& -1& 0 &1& 2\\
\hline y= 3^x & 0 & \frac{1}{3}& 1 & 3 & 9 \\[2pt]
\hline
\end{array} \)

Corresponding coordinate points for the transformation \( g(x) = -2 \cdot 3^{x-5}+6 \) would be

\( \begin{array}{|r|c|c|c|c| c |}
\hline\text{New }x \text{ is: } \:\;\quad x+5& \text{HA}& 4& 5 &6& 7\\
\hline\text{New } y \text{ is: } -2y+6 & 6 & 5\frac{1}{3}& 4 & 0 & -12 \\[2pt]
\hline
\end{array} \)

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:

1. Shift left 1 unit ( \( x \rightarrowx-1 \) )
2. Reflect over the \(y\)-axis ( \( x \rightarrow-x\) )
3. Vertically stretch by a factor of 3 ( \( y \rightarrow3y\) )
4. Shift down \(4\) units ( \( y \rightarrowy-4\) )

If instead we re-wrote the function in the form: \( f(x) = 3 \cdot 2^{-(x-1)}-4 \), the transformations would be

1. Reflect over the \(y\)-axis ( \( x \rightarrow-x\) )
2. Shift right1 unit ( \( x \rightarrowx+1 \) )
3. Vertically stretch by a factor of 3 ( \( y \rightarrow3y\) )
4. Shift down \(4\) units ( \( y \rightarrowy-4\) )

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

\( \begin{array}{|r|c|c|c|c| c |}
\hline\text{Parent function: }x& \text{HA}& -1& 0 &1& 2\\
\hline y= 2^x & 0 & \frac{1}{2}& 1 & 2 & 4 \\[2pt]
\hline
\end{array} \)

Corresponding coordinate points for the transformation \( f(x) = 3 \cdot 2^{-x+1}-4 \) would be

\( \begin{array}{|r|c|c|c|c| c |}
\hline\text{New } x \text{ is: } -x+1& \text{HA}& 2& 1 &0& -1\\
\hline\text{New } y \text{ is: } 3y-4& -4 & -2\frac{1}{2}& -1 & 2 & 8\\[2pt]
\hline
\end{array} \)

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:

1. Shiftleft \(4\)units ( \( x \rightarrowx-4\) )
2. Horizontally stretch by a factor of 2 ( \( x \rightarrow2x\) )
3. Shift down \(3\) units ( \( y \rightarrowy-3\) )

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

\( \begin{array}{|r|c|c|c|c| c |}
\hline\text{Parent function: }x& \text{HA}& -1& 0 &1& 2\\
\hline y= 2^x & 0 & \frac{1}{2}& 1 & 2 & 4 \\[2pt]
\hline
\end{array} \)

Corresponding coordinate points for the transformation would be

\( \begin{array}{|r|c|c|c|c| c |}
\hline\text{New }x \text{ is: } 2(x-4)& \text{HA}& -10& -8 &-6& -4\\
\hline\text{New } y \text{ is: } y-3& -3 & -2.5& -2 & -1 & 1\\[2pt]
\hline
\end{array} \)

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

1. Horizontally stretch by a factor of 2 ( \( x \rightarrow2x\) )
2. Shiftleft \(8\)units ( \( x \rightarrowx-8\) )
3. Shift down \(3\) units ( \( y \rightarrowy-3\) )

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.