Download Review Sheet for Graphs of Exponential and Logarithmic Functions | MAC 1105 and more Study notes Algebra in PDF only on Docsity!
MAC 1105 Handout # 4
Graphs of Exponential and Logarithmic Functions
In this handout, you will familiarize with the graphs of Exponential and Logarithmic functions.
The Exponential Function
The exponential function with base b is defined by f ( ) x = abx. Since f (0)= ab^0 = a , a is
the y -intercept of the function. If the base is greater than 1 ( b >1), the function increases (exponential growth). If the base is between 0 and 1 (0< b <1), the function decreases
(exponential decay). The domain of the exponential function f ( ) x = abx is all the real
numbers x. The range of the exponential function is all the values of y ≥ 0.
Example 1: Compare the graphs of f ( ) x = 2 x and f ( ) x = 10 x
Go to and type in the equations Y 1 (^) = 2 ^ x and Y 2 (^) = 10 ^ x, and select 4
( ZDecimal ). If we change the styles of the curves, we obtain the graph picture, where the graph in bold is the function Y 2 (^) = 10 ^ x.
Notice that both functions are of the form f ( ) x = abx since Y 1 (^) = 2 ^ x with a =1 and b = 2,
and Y 2 (^) = 10 ^ x with a =1 and b =10.
If we compare the two graphs, we see that the functions are increasing since the base b > for both of them. Since a =1 for both functions, their y -intercept = 1.
f ( ) x = 10 x grows faster than f ( ) x = 2 x as the positive values of x increase, and f ( ) x = 10 x decays faster than f ( ) x = 2 x as the negative values of x decrease. We can say
that the larger the base b , the faster the function grows.
Example 2: Compare the graphs of f ( ) x = 2 3⋅ x and f ( ) x = 2 x
Go to and type in the equations Y 1 (^) = 2*3 ^ x and Y 2 (^) = 2 ^ x and select 4
( ZDecimal). If we change the styles of the curves, we obtain the picture below, where the graph in bold is the function Y 2 (^) = 2 ^ x.
Since a = 2 for Y 1 , and a =1 for Y 2 , the y -intercept of Y 1 is 2 while the y -intercept of Y 2 (the bold curve) is 1.
Example 3: Compare the graphs of f ( ) x = 2 x and f ( ) x = ex
Go to and type in the equations Y 1 = 2^( x ) and Y 2 = e^( x ) , and select 4 ( ZDecimal ). e ^( x ) can be obtained by pressing ` and the L keys. If we change the styles of the curves, we obtain the picture below, where the graph in bold is the function Y 2 = ex.
If we compare the two graphs we see that both graphs cross the y -axis at y = 1. Since e ≈ 2.718is larger than 2, we see that f ( ) x = ex grows faster than f ( ) x = 2 x. All
the exponential functions with base b >1 will have the same shape. The difference will be that the higher the value of the base, the faster the function will increase.
Example 4: Compare the graphs of f ( ) x = ex and f ( ) x = e − x
Go to and type in the equations Y 1 = e^( x ) and Y 2 = e^(- x ) , and select 4 ( ZDecimal ). e ^( x ) can be obtained by pressing ` and the L keys. To enter e^(- x ) , choose the _ key (to the left of e). If we change the styles of the curves, we obtain
the picture below, where the graph in bold is the function Y 2 = e − x.
Graphical Solution of Exponential and Logarithmic Equations
Example 7: Find the solution of the equation 2 x^ +^1 = 1 − x graphically by using the intersection feature of your calculator. Round to three decimal places. (Note: Should you need a review of the “intersection feature,” please review handout #1.)
Let Y1 = 2 x +1^ , and Y2 = 1- x. Select the window x : [-8, 8] and y:[-5, 5] shown below.
By pressing ` $, and choosing option 5 (Intersect), selecting the two curves and
a guess near the intersection, we obtain the point of intersection. The solution to three decimal places is x = -.457. This equation cannot be solved by algebraic methods.
Example 8: Find the solution of the equation ln( x + 1) − 1 + x = 0 graphically.
Let Y1 = ln( x + 1) − 1 + x , and Y2 = 0. Select the window x : [-8, 8] and y:[-5, 5] shown
below. By pressing ` $, , choosing option 5 (Intersect), and the curves and a
guess, we obtain the solution. The solution to three decimal places is x = .557. Notice that solving the equation ln( x + 1) − 1 + x = 0 is the same as finding the x-intercepts (or zeros)
of the graph of y = ln( x + 1) − 1 + x.
Example 9: Find the solution to the equation log ( ) 2 x + x = 0 graphically. Round to three
decimal places
By using the change of base formula, let Y1 = log( ) / log(2) x + x. Select 4
( ZDecimal ) window. By pressing ` $, , choosing option 2 (Zero), and
selecting a left bound, a right bound and a guess, we obtain the zero ( x -intercept). The solution to the equation to three decimal places is x =.
Example 10: Find the maximum value of y = xe − x graphically. Round to three decimal
places Let Y1 = xe ^ ( − x ). Select 4 ( ZDecimal ) window
To find the relative maximum, we use CALC. Press ` $, , and select option 4. Move the cursor keys to find a left bound (a point to the left of the desired minimum), a right bound (a point to the right of the desired minimum) and a guess (any point between the two arrow heads). That will give you the relative maximum point (1.000, 0.368). (Note: Should you need further review finding “maximum and minimum values” with the grapher,” please review handout #3.)
Note: Some calculators will give x = 1 instead of 1.0000004. Other calculators will give .9999999. It all depends on how the calculator rounds the numbers.
End of Handout # 4
