5.1 Quadratic functions  (Page 6/15)

We begin by solving for when the output will be zero.

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

We know that $a=2.$ Then we solve for $h$ and $k.$

So now we can rewrite in standard form.

We can now solve for when the output will be zero.

The graph has x -intercepts at $(\mathrm{-1}-\sqrt{3},0)$ and $(\mathrm{-1}+\sqrt{3},0).$

We can check our work by graphing the given function on a graphing utility and observing the $x\text{-}$ intercepts. See [link] .

1. The ball reaches the maximum height at the vertex of the parabola.
   $$\begin{array}{ccc}\hfill h& =& -\frac{80}{2(\mathrm{-16})}\hfill \\ & =& \frac{80}{32}\hfill \\ & =& \frac{5}{2}\hfill \\ & =& 2.5\hfill \end{array}$$

   The ball reaches a maximum height after 2.5 seconds.

2. To find the maximum height, find the $y\text{-}$ coordinate of the vertex of the parabola.
   $$\begin{array}{ccc}\hfill k& =& H\left(-\frac{b}{2a}\right)\hfill \\ & =& H\left(2.5\right)\hfill \\ & =& \mathrm{-16}{\left(2.5\right)}^2+80\left(2.5\right)+40\hfill \\ & =& 140\hfill \end{array}$$

   The ball reaches a maximum height of 140 feet.

3. To find when the ball hits the ground, we need to determine when the height is zero, $H\left(t\right)=0.$

   We use the quadratic formula.

   $$\begin{array}{ccc}\hfill t& =& \frac{\mathrm{-80}\pm \sqrt{{80}^2-4(\mathrm{-16})(40)}}{2(\mathrm{-16})}\hfill \\ & =& \frac{\mathrm{-80}\pm \sqrt{8960}}{\mathrm{-32}}\hfill \end{array}$$

   Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

   $\begin{array}{l}\hfill \\ \hfill \\ \begin{array}{lll}t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458\hfill & \text{or}\hfill & t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458\hfill \end{array}\hfill \end{array}$

   The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See [link] .

   

   Note that the graph does not represent the physical path of the ball upward and downward. Keep the quantities on each axis in mind while interpreting the graph.