3.2 Quadratic functions  (Page 6/14)

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

Let’s begin by writing the quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

When applying the quadratic formula , we identify the coefficients $a,b\text{ and }c.$ For the equation $x^2+x+2=0,$ we have $a=1, b=1, \text{and} c=2.$ Substituting these values into the formula we have:

The solutions to the equation are $x=\frac{-1+i\sqrt{7}}{2}$ and $x=\frac{-1-i\sqrt{7}}{2}$ or $x=\frac{-1}{2}+\frac{i\sqrt{7}}{2}$ and $x=\frac{-1}{2}-\frac{i\sqrt{7}}{2}.$

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

   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.
   $$\begin{array}{l}k=H\left(-\frac{b}{2a}\right)\hfill \\ =H\left(2.5\right)\hfill \\ =\mathrm{-16}{\left(2.5\right)}^2+80\left(2.5\right)+40\hfill \\ =140\hfill \end{array}$$

   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.

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

   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]