Chapter 8.4, Problem 64E

To determine

To find:The quadratic function that models the path of the flare with the maximum height of 300 meters.

Answer to Problem 64E

The quadratic equation is

  $f\left(x\right)=-\frac{1}{12}{\left(x-59\right)}^2+300$

Explanation of Solution

Given information:

A flare is launched from the boat and travels in the parabolic path until reaching the water. The vertex is $\left(h,k\right)\equiv \left(59,300\right)$

Formula used:

  $f\left(x\right)=a{\left(x-h\right)}^2+k$

Calculation:

Let $f\left(x\right)=a{\left(x-h\right)}^2+k$

As $h=59$ and $k=300$ we have $f\left(x\right)=a{\left(x-59\right)}^2+300$

As (119, 0) lies on the function, putting the values in function, we get

  $\begin{array}{l}0=a{\left(119-59\right)}^2+300\\ 0=3600a+300\\ a=-\frac{300}{3600}\\ a=-\frac{1}{12}\end{array}$

Thus, quadratic equation is

  $f\left(x\right)=-\frac{1}{12}{\left(x-59\right)}^2+300$

Plotting $f\left(x\right)$ :

Graph the axis of symmetry: Since $h=59$ axis of symmetry is $x=59$

Plot the vertex: Vertex is $\left(h,k\right)\equiv \left(59,300\right)$

Plot 2 more coordinates. We are given that $\left(119,0\right)$ lies on graph. Finding y-coordinate for $x=29$ , we get that $\left(29,225\right)$ lies on the graph.

Draw smooth curve through the points.

  

Conclusion:

The quadratic equation is

  $f\left(x\right)=-\frac{1}{12}{\left(x-59\right)}^2+300$