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.
- 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: