## Solving Maximum and Minimum Applications

Contents

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.

- How many seconds will it take the volleyball to reach its maximum height?
- 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.

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

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: