# Solving Maximum and Minimum Applications

## Solving Maximum and Minimum Applications

Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic equation. The y-coordinate of the vertex is the minimum y-value of a parabola that opens upward. It is the maximum y-value of a parabola that opens downward. See the figure below.

### Minimum or Maximum Values of a Quadratic Equation

The y-coordinate of the vertex of the graph of a quadratic equation is the

• minimum value of the quadratic equation if the parabola opens upward.
• maximum value of the quadratic equation if the parabola opens downward.

## Example

Find the minimum value of the quadratic equation $$y={x}^{2}+2x-8$$.

### Solution

 Since a is positive, the parabola opens upward. The quadratic equation has a minimum. Find the axis of symmetry. The axis of symmetry is $$x=-1$$. The vertex is on the line $$x=-1.$$ Find y when $$x=-1.$$ The vertex is $$\left(-1,-9\right)$$. Since the parabola has a minimum, the y-coordinate of the vertex is the minimum y-value of the quadratic equation. The minimum value of the quadratic is $$-9$$ and it occurs when $$x=-1$$. Show the graph to verify the result.

We have used the formula

$$h=-16{t}^{2}+{v}_{0}t+{h}_{0}$$

to calculate the height in feet, $$h$$, of an object shot upwards into the air with initial velocity, $${v}_{0}$$, after $$t$$ seconds.

This formula is a quadratic equation in the variable $$t$$, so its graph is a parabola. By solving for the coordinates of the vertex, we can find how long it will take the object to reach its maximum height. Then, we can calculate the maximum height.

## Example

The quadratic equation $$h=-16{t}^{2}+{v}_{0}t+{h}_{0}$$ models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.

1. How many seconds will it take the volleyball to reach its maximum height?
2. Find the maximum height of the volleyball.

### Solution

$$h=-16{t}^{2}+176t+4$$

Since a is negative, the parabola opens downward.

The quadratic equation has a maximum.

1.

$$\begin{array}{cccc}\text{Find the axis of symmetry.}\hfill & & & \phantom{\rule{4em}{0ex}}\begin{array}{c}t=-\frac{b}{2a}\hfill \\ t=-\frac{176}{2\left(-16\right)}\hfill \\ t=5.5\hfill \end{array}\hfill \\ & & & \phantom{\rule{4em}{0ex}}\text{The axis of symmetry is}\phantom{\rule{0.2em}{0ex}}t=5.5.\hfill \\ \text{The vertex is on the line}\phantom{\rule{0.2em}{0ex}}t=5.5.\hfill & & & \phantom{\rule{4em}{0ex}}\text{The maximum occurs when}\phantom{\rule{0.2em}{0ex}}t=5.5\phantom{\rule{0.2em}{0ex}}\text{seconds.}\hfill \end{array}$$

2.
 Find h when $$t=5.5$$. Use a calculator to simplify. The vertex is $$\left(5.5,488\right)$$. Since the parabola has a maximum, the h-coordinate of the vertex is the maximum y-value of the quadratic equation. The maximum value of the quadratic is 488 feet and it occurs when $$t=5.5$$ seconds.

## Optional Video

### Resources:

You can access these resources for additional instruction and practice graphing quadratic equations: