Question

This picture is the paragraph of information to answer the questions. The second picture is the questions

Answer 1

Answers:

Question 1

a) It could not represent the scenario because the vertex of the parabola is located at (-5,9), and x=-5 is a region beyond the bank.

b) This function does not fit the scenario because the parabola opens up, as if the fish fell towards the sky.

c) This function does not fit the scenario because the expression is negative for all values of x, which means that the fish always remains under water.

Question 2

The x-value at the center of the boat is 5.

Question 3

The fish jumps 4 feet high.

Question 4

The zeros of the function are x=3 and x=7 and they represent the locations over the x-axis where the fish comes out of the water and re-enters the river.

Question 5

The fish comes out of the water 2 feet away from the center of the boat.

Question 6

The domain is: 0≤x≤50.

The range is: -2021≤y≤4.

Question 7

a) The path of the fish is increasing at the interval [0,5).

b) The path of the fish is decreasing at the interval (5,50]

Question 8

a) The fish swims under water at the intervals [0,3) and (7,50].

b) The fish swims abov the water at the interval (3,7).

Step-by-step explanation:

Question 1:

Write the functions in vertex form.

a)

`\begin{gathered} F\left(x\right)=-x^2-10x-16 \\ =-\left(x+5\right)^2+9 \end{gathered}`

The vertex is located at (-5,9), but the bank is the line x=0 and the region x<0 corresponds to land. The fish cannot swim on the land, so it would be impossible for the fish to reach that point. Additionally, the function is negative for all positive values of x, so in the region that corresponds to the river, the fish never comes out of the water.

b)

`F\left(x\right)=x^2-6x+13`

Since the coefficient of the quadratic term is positive, the parabola opens up. Then, this trajectory corresponds to an object that "falls upwards", which is not possible.

c)

`\begin{gathered} F\left(x\right)=-x^2+6x-13 \\ =-\left(x-3\right)^2-4 \end{gathered}`

Notice that the function is negative for all values of x, which can be interpreted as if the fish never came out of the water, which does not correspond to the described situation.

Question 2:

Write the expression that describes the trajectory of the fish in vertex form. The x-coordinate of the vertex corresponds to the center of the boat.

`\begin{gathered} F\left(x\right)=-x^2+10x-21 \\ =-\left(x-5\right)^2+4 \end{gathered}`

The vertex of the parabola is (5,4), so the center of the boat is located at x=5.

Question 3:

The maximum height of the fish corresponds to the y-coordinate of the vertex. So, the fish's maximum height is y=4 (4 feet).

Question 4:

Set F(x)=0 and solve for x:

`\begin{gathered} F\left(x\right)=0 \\ \Rightarrow-\left(x-5\right)^2+4=0 \\ \Rightarrow4=\left(x-5\right)^2 \\ \Rightarrow\left(x-5\right)^2=4 \\ \Rightarrow x-5=±√(4) \\ \Rightarrow x-5=±2 \\ \Rightarrow x=5±2 \\ \therefore x_1=3,x_2=7 \\ \end{gathered}`

Then, the zeros of the function are x=3 and x=7, they represent the horizontal location at which the fish is located at the surface of the river, so they are the points where the fish comes out of the water and re-enters the water.

Question 5:

The points x=3 and x=7 are both 2 units away from x=5. Then, the fih comes out of the water and re-enters the water 2 feet away from the center of the boat.

Question 6:

a)

The domain corresponds to all the values of x that the function can take. Since the river is 50 feet wide and the bank is the y-axis, then, the river covers the region 0≤x≤50, which is the same as the domain.

b)

The range corresponds to all the values over the y-axis that the function can take when evaluated at values from the domain.

To find the range, we have to find the maximum and minimum values of F(x) for 0≤x≤50. Since the maximum value of the function is 4 (the maximum height), just check for the values of F(0) and F(50) to find the possibilities for the minimum:

`\begin{gathered} F\left(0\right)=-\left(0-5\right)^2+4 \\ =-25+4 \\ =-21 \end{gathered}`
`\begin{gathered} F\left(50\right)=-\left(50-5\right)^2+4 \\ =-\left(45\right)^2+4 \\ =-2025+4 \\ =-2021 \end{gathered}`

Then, the minimum value of the function F for the values in the domain is -2021. Therefore, the range is: -2021≤y≤4.

Question 7:

The path of the fish increases until it reaches it maximum point at x=5, then it decreases. Then:

a) The path of the fish increases at 0≤x<5.

b) The path of the fish decreases as 5.

Question 8:

The fish is initially under water, it comes out at the first zero of the function x=3, it travels through the air until it re-enters the water at x=7 and it continues under water from that point on. Then:

a) The fish is under water in the interval 0≤x<3 and 7.

b) The fish is above the water in the interval 3.