1. Chapter 7 Section 1

3. EXAMPLE 1 Graph y = b for b > 1 x SOLUTION Make a table of values.STEP 1 STEP 2 Plot the points from the table. Graph y =. x 2 STEP 3 Draw, from left to right, a smooth curve that begins just above the x -axis, passes through the plotted points, and moves up to the right.

4. EXAMPLE 2 Graph y = ab for b > 1 x Graph the function. a. y = 4 1 2 x SOLUTION Plot and (1, 2).Then, from left to right, draw a curve that begins just above the x -axis, passes through the two points, and moves up to the right. 0, 1 2 a.

5. EXAMPLE 2 Graph the function. Graph y = ab for b > 1 x b. y = – 5 2 x Plot (0, –1) and. Then,from left to right, draw a curve that begins just below the x -axis, passes through the two points,and moves down to the right. b.b. 1, – 5 2 SOLUTION

6. EXAMPLE 3 Graph y = ab + k for b > 1 x–hx–h Graph y = 4 2 – 3. State the domain and range. x – 1 SOLUTION Begin by sketching the graph of y = 4 2, which passes through (0, 4) and (1, 8). Then translate the graph right 1 unit and down 3 units to obtain the graph of y = 4 2 – 3.The graph’s asymptote is the line y = –3. The domain is all real numbers, and the range is y > –3. x x – 1

7. GUIDED PRACTICE for Examples 1, 2 and 3 Graph the function. State the domain and range. 1. y = 4 x

8. GUIDED PRACTICE for Examples 1, 2 and 3 Graph the function. State the domain and range. 2. y = 3 2 3 x

9. GUIDED PRACTICE for Examples 1, 2 and 3 Graph the function. State the domain and range. 3. f (x) = 3 + 2 x + 1

11. EXAMPLE 4 Solve a multi-step problem Write an exponential growth model giving the number n of incidents t years after 1996. About how many incidents were there in 2003 ? In 1996, there were 2573 computer viruses and other computer security incidents. During the next 7 years, the number of incidents increased by about 92% each year. Computers

12. EXAMPLE 4 Solve a multi-step problem Graph the model. Use the graph to estimate the year when there were about 125,000 computer security incidents. SOLUTION STEP 1 The initial amount is a = 2573 and the percent increase is r = 0.92. So, the exponential growth model is: n = a (1 + r) t = 2573(1 + 0.92) t = 2573(1.92) t Write exponential growth model. Substitute 2573 for a and 0.92 for r. Simplify.

13. EXAMPLE 4 Solve a multi-step problem Using this model, you can estimate the number of incidents in 2003 (t = 7) to be n = 2573(1.92) 247,485. 7 STEP 2 The graph passes through the points (0, 2573) and (1,4940.16). Plot a few other points. Then draw a smooth curve through the points.

14. EXAMPLE 4 Solve a multi-step problem STEP 3 Using the graph, you can estimate that the number of incidents was about 125,000 during 2002 (t 6).

15. GUIDED PRACTICE for Example 4 4. What If? In Example 4, estimate the year in which there were about 250,000 computer security incidents. SOLUTION 2003

16. GUIDED PRACTICE for Example 4 5. In the exponential growth model y = 527(1.39), identify the initial amount,the growth factor, and the percent increase. x SOLUTION Initial amount: 527 Growth factor 1.39 Percent increase 39%

18. EXAMPLE 5 Find the balance in an account You deposit $4000 in an account that pays 2.92% annual interest. Find the balance after 1 year if the interest is compounded with the given frequency. FINANCE a. Quarterly b. Daily

19. EXAMPLE 5 Find the balance in an account = 4000 1 + 0.0292 4 4 1 = 4000(1.0073) 4 = 4118.09 P = 4000, r = 0.0292, n = 4, t = 1 Simplify. Use a calculator. ANSWER The balance at the end of 1 year is $4118.09. SOLUTION a. With interest compounded quarterly, the balance after 1 year is: A = P 1 + r n nt Write compound interest formula.

20. EXAMPLE 5 Find the balance in an account b. With interest compounded daily, the balance after 1 year is: A = P 1 + r n nt = 4000 1 + 0.0292 365 365 1 = 4000(1.00008) 365 = 4118.52 Write compound interest formula. P = 4000, r = 0.0292, n = 365, t = 1 Simplify. Use a calculator. ANSWER The balance at the end of 1 year is $4118.52.

21. GUIDED PRACTICE for Example 5 6. FINANCE You deposit $2000 in an account that pays 4% annual interest. Find the balance after 3 years if the interest is compounded daily. $2254.98 ANSWER a. With interest compounded daily, the balance after 3 years is: