1. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 1 Chapter 4 Exponential Functions

2. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 2 4.4 Finding Equations of Exponential Functions

3. 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.

4. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 4 Solution For f(x) = ab x, 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

5. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 5 Solution Check the result with a graphing calculator.

6. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 6 Solving Equations of the Form b n = k for b To solve an equation of the form b n = 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.

7. 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. 5.42b 6 – 3.19 = 43.742.

8. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 8 Solution 1.

9. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 9 Solution 2.

10. 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 = ab x of the exponential curve that contains the points (0, 3) and (4, 70). Round the value of b to two decimal places.

11. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 11 Solution Since the y-intercept is (0, 3), the equation has the form y = 3b x. Next, substitute (4, 70) in the equation and solve for b:

12. 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.

13. 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.

14. 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 = ab x of the exponential curve that contains (2, 5) and (5, 63). Round the values of a and b to two decimal places.

15. 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 = ab x, we have the following system of equations:

16. 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:

17. 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:

18. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 18 Solution We can now solve for b by finding the cube root:

19. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 4.4, Slide 19 Solution Substitute 2.33 for the constant b in the equation y = ab x : To find a, substitute the coordinates of the given point (2, 5) into the equation:

20. 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.

21. 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.