Upload
barrie-charles
View
218
Download
0
Embed Size (px)
DESCRIPTION
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 3 Example: Finding an Equation of an Exponential Curve An exponential curve contains the points listed in the table below. Find an equation of the curve.
Citation preview
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1
Chapter 4Exponential Functions
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 2
4.4 Finding Equations of Exponential Functions
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 3
Example: Finding an Equation of an Exponential Curve
An exponential curve contains the points listed in the table below. Find an equation of the curve.
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 4
Solution
For f(x) = abx, recall that the y-intercept is (0, a). We see from the table that the y-intercept is (0, 3), so a = 3. As the value of x increases by 1, the value of y is multiplied by 2. By the base multiplier property, b = 2. Therefore, and equation of the curve is
f(x) = 3(2)x
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 5
Solution
Check the result with a graphing calculator.
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 6
Solving Equations of the Form bn = k for b
To solve an equation of the form bn = k for b,
1. If n is odd, the real-number solution is
2. If n is even and k ≥ 0, the real number solutions are
3. If n is even and k < 0, there is no real-number solution.
1 .nk
1 .nk
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 7
Example: One-Variable Equations Involving Exponents
Find all real-number solutions. Round any results to the second decimal place.
1. 5.42b6 – 3.19 = 43.74 2.9
4703
bb
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 8
Solution
1. 65.42 3.19 43.74b 65.42 43.74 3.19b 65.42 46.93b 6 46.93
5.42b
1 646.935.42
b 1.43b
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 9
Solution
2.9
4703
bb
5 703
b
1 5703
b
1.88b
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 10
Example: Finding an Equation of an Exponential Curve
Find an approximate equation y = abx of the exponential curve that contains the points (0, 3) and (4, 70). Round the value of b to two decimal places.
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 11
SolutionSince the y-intercept is (0, 3), the equation has the form y = 3bx. Next, substitute (4, 70) in the equation and solve for b:
470 3b43 70b 4 70
3b
1 4703
b
2.20b
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 12
Solution
So, our equation is y = 3(2.20)x; its graph contains the given point (0, 3). Since we rounded the value of b, the graph of the equation comes close to, but does not pass through, the given point (4, 70). We use a graphing calculator to verify our work.
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 13
Dividing Left Sides and Right Sides of Two Equations
If a = b, c = d, c ≠ 0, and d ≠ 0, then
In words, the quotient of the left sides of two equations is equal to the quotient of the right sides.
a bc d
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 14
Example: Finding the Equation of an Exponential Curve
Find an approximate equation y = abx of the exponential curve that contains (2, 5) and (5, 63). Round the values of a and b to two decimal places.
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 15
Solution
Since both of the ordered pairs (2, 5) and (5, 63) must satisfy the equation y = abx, we have the following system of equations:
25 ab563 ab
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 16
Solution
It will be slightly easier to solve this system if we switch the equations to list the equation with the greater exponent of b first:
25 ab
563 ab
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 17
Solution
We divide the left sides and divide the right sides of the two equations to get the following result for nonzero a and b:
5
2635
baab
3635
b
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 18
Solution
We can now solve for b by finding the cube root:
3 635
b
1 3635
b
2.33b
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 19
Solution
Substitute 2.33 for the constant b in the equation y = abx:
To find a, substitute the coordinates of the given point (2, 5) into the equation:
2.33 xy a
25 2.33a
25
2.33a
0.92a
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 20
Solution
So, an equation that approximates the exponential curve that passes through (2, 5) and (5, 63) is y = 0.92(2.33)x.
We use a graphing calculator to verify our work.
Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 21
Exponential Function
We can find an equation of an exponential function by using the base multiplier property or by using two points. Both methods give the same result.