Question

unit 5 part 2 test review2
5a. graph y = 4·3^x y = 5·(1/3)^x y = 8·(0.9)^x y = 3·(-1/2)^x

1. find the exponential function for the table:

several x - y value tables are shown
8a. a population of cells is growing in a science lab. the scenario is modeled by the exponential function y = 7250(3)^x. what does the 7250 represent in real - world context?
8b. a population of fish is growing in an aquarium. the scenario is modeled by the exponential function y = 4600(9)^x. what does the 4600 represent in real - world context?
8c. a population of bears is growing on an alaskan island. the scenario is modeled by the exponential function y = 3(5)^x. what does the 3 represent in real - world context?
9a. the function for an investment is as follows: y = 500(1.2)^x. how much will the investment be worth in 5 years? round to 2 decimal places
9a. the function for an investment is as follows: y = 3000(1.7)^x. how much will the investment be worth in 7 years? round to 2 decimal places
9a. the function for an investment is as follows: y = 21000(1.45)^x. how much will the investment be worth in 10 years? round to 2 decimal places

Explanation:

7.

The general form of an exponential function is $y = ab^{x}$. When $x = 0$, $y=a$.

• For the first table:
• When $x = 0$, $y = 3$, so $a = 3$. When $x=1$, $y = 15$. Substitute into $y=ab^{x}$, we get $15=3b^{1}$, then $b = 5$. The exponential - function is $y = 3\cdot5^{x}$.
• For the second table: When $x = 0$, $y = 4$, so $a = 4$. When $x = 1$, $y=12$. Substitute into $y = ab^{x}$, we get $12 = 4b^{1}$, then $b = 3$. The exponential function is $y=4\cdot3^{x}$.
• For the third table: When $x = 0$, $y=0.08$, so $a = 0.08$. When $x = 1$, $y = 0.8$. Substitute into $y=ab^{x}$, we get $0.8=0.08b^{1}$, then $b = 10$. The exponential function is $y = 0.08\cdot10^{x}$.
• For the fourth table: When $x = 0$, $y=-4$, so $a=-4$. When $x = 1$, $y = 8$. Substitute into $y=ab^{x}$, we get $8=-4b^{1}$, then $b=-2$. The exponential function is $y=-4\cdot(-2)^{x}$.
• For the fifth table: When $x = 0$, $y=\frac{5}{3}$, so $a=\frac{5}{3}$. When $x = 1$, $y=\frac{5}{3}$. Substitute into $y=ab^{x}$, we get $\frac{5}{3}=\frac{5}{3}b^{1}$, then $b = 1$. The exponential function is $y=\frac{5}{3}\cdot1^{x}=\frac{5}{3}$ (but this is a constant function, not a typical exponential growth/decay function. If we assume an error and use the ratio of $y$ - values, from $x = 1$ to $x = 2$, $\frac{y(2)}{y(1)}=\frac{\frac{10}{9}}{\frac{5}{3}}=\frac{2}{3}$, so the function is $y=\frac{5}{3}\cdot(\frac{2}{3})^{x}$).
• For the sixth table: When $x = 0$, $y=-6$, so $a=-6$. When $x = 1$, $y=-24$. Substitute into $y=ab^{x}$, we get $-24=-6b^{1}$, then $b = 4$. The exponential function is $y=-6\cdot4^{x}$.

8a.

In the exponential - growth model $y = ab^{x}$, when $x = 0$, $y=a$. In the function $y = 7250(3)^{x}$, the value $7250$ represents the initial population of cells in the science lab (the population of cells at time $x = 0$).

8b.

In the exponential - growth model $y = ab^{x}$, for the function $y = 4600(9)^{x}$, the value $4600$ represents the initial population of fish in the aquarium (the population of fish at time $x = 0$).

8c.

In the exponential - growth model $y = ab^{x}$, for the function $y = 3(5)^{x}$, the value $3$ represents the initial population of bears on the Alaskan island (the population of bears at time $x = 0$).

9a (first part: $y = 500(1.2)^{x}$, $x = 5$)

Substitute $x = 5$ into the function $y = 500(1.2)^{x}$:

• Step 1: Calculate $(1.2)^{5}$

$(1.2)^{5}=1.2\times1.2\times1.2\times1.2\times1.2 = 2.48832$.

• Step 2: Multiply by 500

$y=500\times2.48832 = 1244.16$.

9a (second part: $y = 3000(1.7)^{x}$, $x = 7$)

Substitute $x = 7$ into the function $y = 3000(1.7)^{x}$:

• Step 1: Calculate $(1.7)^{7}$

$(1.7)^{7}=1.7\times1.7\times1.7\times1.7\times1.7\times1.7\times1.7\approx24.137565$.

• Step 2: Multiply by 3000

$y = 3000\times24.137565=72412.695\approx72412.70$.

9a (third part: $y = 21000(1.45)^{x}$, $x = 10$)

Substitute $x = 10$ into the function $y = 21000(1.45)^{x}$:

• Step 1: Calculate $(1.45)^{10}$

$(1.45)^{10}\approx53.794769$.

• Step 2: Multiply by 21000

$y=21000\times53.794769 = 1129690.149\approx1129690.15$.

Answer:

7.

The general form of an exponential function is $y = ab^{x}$. When $x = 0$, $y=a$.

• For the first table:
• When $x = 0$, $y = 3$, so $a = 3$. When $x=1$, $y = 15$. Substitute into $y=ab^{x}$, we get $15=3b^{1}$, then $b = 5$. The exponential - function is $y = 3\cdot5^{x}$.
• For the second table: When $x = 0$, $y = 4$, so $a = 4$. When $x = 1$, $y=12$. Substitute into $y = ab^{x}$, we get $12 = 4b^{1}$, then $b = 3$. The exponential function is $y=4\cdot3^{x}$.
• For the third table: When $x = 0$, $y=0.08$, so $a = 0.08$. When $x = 1$, $y = 0.8$. Substitute into $y=ab^{x}$, we get $0.8=0.08b^{1}$, then $b = 10$. The exponential function is $y = 0.08\cdot10^{x}$.
• For the fourth table: When $x = 0$, $y=-4$, so $a=-4$. When $x = 1$, $y = 8$. Substitute into $y=ab^{x}$, we get $8=-4b^{1}$, then $b=-2$. The exponential function is $y=-4\cdot(-2)^{x}$.
• For the fifth table: When $x = 0$, $y=\frac{5}{3}$, so $a=\frac{5}{3}$. When $x = 1$, $y=\frac{5}{3}$. Substitute into $y=ab^{x}$, we get $\frac{5}{3}=\frac{5}{3}b^{1}$, then $b = 1$. The exponential function is $y=\frac{5}{3}\cdot1^{x}=\frac{5}{3}$ (but this is a constant function, not a typical exponential growth/decay function. If we assume an error and use the ratio of $y$ - values, from $x = 1$ to $x = 2$, $\frac{y(2)}{y(1)}=\frac{\frac{10}{9}}{\frac{5}{3}}=\frac{2}{3}$, so the function is $y=\frac{5}{3}\cdot(\frac{2}{3})^{x}$).
• For the sixth table: When $x = 0$, $y=-6$, so $a=-6$. When $x = 1$, $y=-24$. Substitute into $y=ab^{x}$, we get $-24=-6b^{1}$, then $b = 4$. The exponential function is $y=-6\cdot4^{x}$.

8a.

In the exponential - growth model $y = ab^{x}$, when $x = 0$, $y=a$. In the function $y = 7250(3)^{x}$, the value $7250$ represents the initial population of cells in the science lab (the population of cells at time $x = 0$).

8b.

In the exponential - growth model $y = ab^{x}$, for the function $y = 4600(9)^{x}$, the value $4600$ represents the initial population of fish in the aquarium (the population of fish at time $x = 0$).

8c.

In the exponential - growth model $y = ab^{x}$, for the function $y = 3(5)^{x}$, the value $3$ represents the initial population of bears on the Alaskan island (the population of bears at time $x = 0$).

9a (first part: $y = 500(1.2)^{x}$, $x = 5$)

Substitute $x = 5$ into the function $y = 500(1.2)^{x}$:

• Step 1: Calculate $(1.2)^{5}$

$(1.2)^{5}=1.2\times1.2\times1.2\times1.2\times1.2 = 2.48832$.

• Step 2: Multiply by 500

$y=500\times2.48832 = 1244.16$.

9a (second part: $y = 3000(1.7)^{x}$, $x = 7$)

Substitute $x = 7$ into the function $y = 3000(1.7)^{x}$:

• Step 1: Calculate $(1.7)^{7}$

$(1.7)^{7}=1.7\times1.7\times1.7\times1.7\times1.7\times1.7\times1.7\approx24.137565$.

• Step 2: Multiply by 3000

$y = 3000\times24.137565=72412.695\approx72412.70$.

9a (third part: $y = 21000(1.45)^{x}$, $x = 10$)

Substitute $x = 10$ into the function $y = 21000(1.45)^{x}$:

• Step 1: Calculate $(1.45)^{10}$

$(1.45)^{10}\approx53.794769$.

• Step 2: Multiply by 21000

$y=21000\times53.794769 = 1129690.149\approx1129690.15$.