Chapter 3, Problem 14CT

To determine

To calculate: The maximum height and distance of the horseshoe.

Answer to Problem 14CT

The required distance is 35.615 feet.

Explanation of Solution

Given information:

The model of the tosses: $y=-0.01x^2+0.3x+2$

Calculation:

The given model of tosses is $y=-0.01x^2+0.3x+2$ .

Rewrite the above equation as:

  $\begin{array}{l}y=-0.01x^2+0.3x+2\\ y=-0.01\left(x^2-30x-200\right)\end{array}$

Simplify the expression $x^2-30x-200$ by using perfect square.

  $\begin{array}{l}y=-0.01\left(x^2-30x-200\right)\\ y=-0.01\left(x^2+30x+225-425\right)\\ y=-0.01\left[{\left(x+15\right)}^2-425\right]\\ y=-0.01{\left(x+15\right)}^2+4.25\end{array}$

The above equation is vertex form.

So, the distance to the vertex is 15 and the maximum height is 4.25.

To find the maximum distance, solve the expression $x^2-30x-200=0$ .

Use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Substitute 1 for a , −30 for b and −200 for c in the above formula and simplify for x .

  $\begin{array}{c}x=\frac{-\left(-30\right)\pm \sqrt{{\left(-30\right)}^2-4\left(1\right)\left(-200\right)}}{2\left(1\right)}\\ =\frac{30\pm \sqrt{900+800}}{2}\\ =\frac{30\pm \sqrt{1700}}{2}\\ =\frac{30\pm 41.23}{2}\end{array}$

Solve the above expression for x .

  $\begin{array}{c}x=\frac{30+41.23}{2}\\ =\frac{71.23}{2}\\ =35.165\end{array}$

And, $\begin{array}{c}x=\frac{30-41.23}{2}\\ =\frac{-11.23}{2}\\ =-5.615\end{array}$

The distance cannot be negative.

Hence, the required distance is 35.615 feet.