Chapter 1.2, Problem 8GP

Solution

What is the maximum height of the football if the football’s path can be modeled by the function $y=-0.025x(x-50)$ ?

The maximum height of the football if the football’s path can be modeled by the function $y=-0.025x(x-50)$ is = $15.625$

Axis of symmetry (x) $=\frac{-b}{2a}=25$

Vertex $=(25,15.625)$

Given:

The given function $y=-0.025x(x-50)$

Concept Used:

• The vertex is the highest point if the parabola opens downward and the lowest point if the parabola opens upward.
• The axis of symmetry is the line that cuts the parabola into 2 matching halves and the vertex lies on the axis of symmetry.

Calculation:

The graph of the given function $y=-0.025x(x-50)$ is -

  

The given function $y=-0.025x(x-50)$

  $\begin{array}{c}y=-0.025x(x-50)\\ y=-0.025x^2-0.025x\cdot (-50)\\ y=-0.025x^2+1.25x\end{array}$

Here $a=-0.025,b=1.25,c=0$

Axis of symmetry (x) $=\frac{-b}{2a}$

  $\begin{array}{l}=\frac{-(1.25)}{2(-0.025)}\\ =\frac{-1.25}{-0.05}\\ =\frac{125}{5}\\ =25\end{array}$

So Axis of symmetry (x) $=\frac{-b}{2a}=25$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

  $\begin{array}{c}y\left(\frac{-b}{2a}\right)=y\left(25\right)\\ y=-0.025x(x-50)\\ y=-0.025\cdot (25)(25-50)\\ =-0.025\cdot (25)(-25)\\ =-0.025\cdot (-625)\\ =15.625\end{array}$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

Vertex $=(25,15.625)$

So, the maximum height of the football if the football’s path can be modeled by the function $y=-0.025x(x-50)$ is = $15.625$