Question

the population of havenport (in tens) can be modeled with the following equation, where t represents the years since 1950. graph the exponential function, rounding to the nearest integer whenever necessary. p(t) = 98(1.05)^t show your work here draw

Explanation:

Step1: Identify the type of function

The function \( p(t) = 98(1.05)^t \) is an exponential function of the form \( y = a(b)^t \), where \( a = 98 \) (initial value), \( b = 1.05 \) (growth factor, \( b>1 \) so it's growth), and \( t \) is the independent variable (years since 1950).

Step2: Find key points (t, p(t))

• When \( t = 0 \) (year 1950):

\( p(0) = 98(1.05)^0 = 98(1) = 98 \). So the point is \( (0, 98) \).

• When \( t = 10 \) (1960):

\( p(10) = 98(1.05)^{10} \approx 98(1.62889) \approx 159.63 \approx 160 \). Point: \( (10, 160) \).

• When \( t = 20 \) (1970):

\( p(20) = 98(1.05)^{20} \approx 98(2.65330) \approx 259.92 \approx 260 \). Point: \( (20, 260) \).

• When \( t = 30 \) (1980):

\( p(30) = 98(1.05)^{30} \approx 98(4.32194) \approx 423.55 \approx 424 \). Point: \( (30, 424) \).

Step3: Plot the points and draw the curve

On the graph (with \( t \) on the x - axis, \( p(t) \) on the y - axis), plot \( (0, 98) \), \( (10, 160) \), \( (20, 260) \), \( (30, 424) \), then draw a smooth exponential growth curve through these points. The curve should start at \( (0, 98) \) and increase, getting steeper as \( t \) increases (since it's exponential growth).

Answer:

To graph \( p(t)=98(1.05)^t \):

1. Plot the initial point \( (0, 98) \) (when \( t = 0 \), population is 98 tens).
2. Calculate and plot additional points (e.g., \( (10, 160) \), \( (20, 260) \), \( (30, 424) \)) using the function.
3. Draw a smooth, increasing exponential curve through these points, starting at \( (0, 98) \) and curving upward (steeper over time).