Exercise 3 Page 75

We first need to consider or find a quadratic function that models a real-life situation. Then, we use it to find the desired information. Suppose two rabbits are introduced to a deserted island. Let P(t) be the number of rabbits after t years. P(t)=- 4t^2+40t+2 Suppose now we want to answer six questions.

1. How many rabbits are there when t=0?
2. After how many years does the population of rabbits reach its maximum?
3. How many rabbits are there when the population reaches its maximum?
4. After how many years will the rabbits will be gone?
5. State a reason for which the population may increase and then decrease.
6. Draw the graph of the function.

Let's work it out!

Question 1

To find how many rabbits there are when t=0, we need to substitute 0 for t in P(t)=- 4t^2+40t+2.

P(t)=- 4t^2+40t+2

Substitute

t= 0

P( 0)=- 4( 0)^2+40( 0)+2

▼

Simplify right-hand side

CalcPow

Calculate power

P(0)=- 4(0)+40(0)+2

Multiply

Multiply

P(0)=0+0+2

AddTerms

Add terms

P(0)=2

This means that when the two rabbits were introduced, they were the only ones in the island. This makes sense, since we are told it was a deserted island.

Question 2

To find out after how many years the population will reach its maximum, we first need to identify a, b, and c. P(t)= - 4t^2+ 40t+ 2 We see above that a= - 4, b= 40, and c= 2. Note that since a= - 4 is negative, the parabola opens downwards. Therefore, to know after how many years the population will reach its maximum, we need to find the first coordinate of the vertex of the parabola. To do so, we will substitute b= 40 and a= - 4 in the expression - b2a.

- b/2a

SubstituteII

a= - 4, b= 40

- 40/2( - 4)

MultPosNeg

a(- b)=- a * b

- 40/- 8

RemoveNegFracAndDenom

- a/- b= a/b

5

We found that the number of rabbits reaches its maximum after 5 years.

Question 3

To calculate how many rabbits there are when the population reaches its maximum, we need to find the second coordinate of the vertex. To do so, we will substitute the first coordinate of the vertex, 5, into the original equation.

P(t)=- 4t^2+40t+2

Substitute

t= 5

P( 5)=- 4( 5)^2+40( 5)+2

▼

Simplify right-hand side

CalcPow

Calculate power

P(5)=- 4(25)+40(5)+2

Multiply

Multiply

P(5)=- 100+200+2

AddTerms

Add terms

P(5)=102

The maximum number of rabbits is 102.

Question 4

The rabbits will be gone when the population is zero. Therefore, we need to find the value of t for which P(t)=0. P(t)=0 ⇔ - 4t^2+40t+2=0 To do so, we will use the Quadratic Formula.

t=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

t=- 40±sqrt(40^2-4( - 4)( 2))/2( - 4)

▼

Simplify right-hand side

CalcPow

Calculate power

t=- 40±sqrt(1600-4(- 4)(2))/2(- 4)

Multiply

Multiply

t=- 40±sqrt(1600+32)/- 8

AddTerms

Add terms

t=- 40±sqrt(1632)/- 8

UseCalc

Use a calculator

lt≈ - 0.0498 t≈ 10

Since t represents time, it cannot be negative. Thus, the rabbits will be gone after approximately 10 years.

Question 5

After being introduced, the rabbits reproduce and the population increases. After 5 years they may start to run out of food or water, and the population starts to decrease until they disappear.

Question 6

To draw the parabola, we will plot the vertex (5,102), the y-intercept (0,2), and its reflection across the axis of symmetry x=5. Then, we will connect the three points with a smooth curve.