Chapter 2.1, Problem 45E

a.

To determine

To Write: An equation of the form $y=a{\left(x-h\right)}^2+k$ with vertex $\left(33,5\right)$ that models the flight path.

Answer to Problem 45E

The equation of the form $y=a{\left(x-h\right)}^2+k$ with vertex $\left(33,5\right)$ that models the flight path is,

  $y=-0.00459{\left(x-33\right)}^2+5$.

Explanation of Solution

Given: The fish leaves the water at $\left(0,0\right)$

Concept Used:

The equation of the form $y=a{\left(x-h\right)}^2+k$ represents a parabola with vertex $\left(h,k\right)$ .

Calculation:

The vertex is $\left(33,5\right)$ , therefore, $h=33\text{ and }k=5$ . Since the fish leaves the tank at $\left(0,0\right)$ , $x=0,y=0$ . Substitute these values in $y=a{\left(x-h\right)}^2+k$ to obtain the value of a.

  $\begin{array}{c}y=a{\left(x-h\right)}^2+k\\ 0=a{\left(0-33\right)}^2+5\\ 0=1089a+5\\ -1089a=5\\ a=-\frac{5}{1089}\\ \approx -0.00459\end{array}$

Substitute the value of a, $h=33\text{ and }k=5$ in $y=a{\left(x-h\right)}^2+k$ to write the equation that represent the path of the fish.

  $y=-0.00459{\left(x-33\right)}^2+5$

Conclusion:

The equation of the form $y=a{\left(x-h\right)}^2+k$ with vertex $\left(33,5\right)$ that models the flight path is,

  $y=-0.00459{\left(x-33\right)}^2+5$

b.

To determine

To Write: The range and domain of the function. What do they represent these values in this model.

Answer to Problem 45E

The range is $\left[0,5\right]$ .

Explanation of Solution

Given: The model that represents the flight of the path is $y=-0.00459{\left(x-33\right)}^2+5$ .

Concept Used:

Domain denotes the distance the fish covers horizontally. Range is the maximum height that the fish can reach.

Calculation:

Calculate the domain by finding the distance covered by the fish horizontally. The distance is calculated by the value of the x-intercepts.

  $y=-0.00459{\left(x-33\right)}^2+5$

Substitute $y=0$ to obtain the values of x,

  $\begin{array}{c}y=-0.00459{\left(x-33\right)}^2+5\\ 0=-0.00459{\left(x-33\right)}^2+5\\ 0.00459{\left(x-33\right)}^2=5\\ {\left(x-33\right)}^2=\frac{5}{0.00459}\\ x-33=\pm \sqrt{\frac{5}{0.00459}}\\ x-33=\pm 33\\ x=\pm 33+33\\ =66,0\end{array}$

Hence, the domain of the function is $\left[0,66\right]$ .

Calculate the range by finding the height covered by the fish. The fish starts from $\left(0,0\right)$ and reaches the vertex $\left(33,5\right)$ . Therefore, the minimum height is 0 and it reaches the maximum height of 5 as the vertex represents the highest point.

Hence, the range is $\left[0,5\right]$ .

Conclusion:

The domain of the function is $\left[0,66\right]$and it denotes the distance the fish covers horizontally.Rangeis $\left[0,5\right]$and it denotes the maximum height that the fish can reach.

c.

To determine

Does the value of a change when the flight has vertex $\left(30,4\right)$ .

Answer to Problem 45E

The value of a is almost same as it was before.

Explanation of Solution

Given: The fish leaves the water at $\left(0,0\right)$ .

Concept Used:

The equation of the form $y=a{\left(x-h\right)}^2+k$ represents a parabola with vertex $\left(h,k\right)$ .

Calculation:

The vertex is $\left(30,4\right)$ , therefore, $h=33\text{ and }k=5$ . Since the fish leaves the tank at $\left(0,0\right)$ , $x=0,y=0$ . Substitute these values in $y=a{\left(x-h\right)}^2+k$ to obtain the value of a.

  $\begin{array}{c}y=a{\left(x-h\right)}^2+k\\ 0=a{\left(0-30\right)}^2+4\\ 0=900a+4\\ -900a=4\\ a=-\frac{4}{900}\\ \approx -0.0044\end{array}$

The value of a shows a negligible change.

Conclusion:

The value of a is almost same as it was before.