Mathematics » Introducing Quadratic Equations » Graphing Quadratic Equations

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.

This figure shows two graphs side by side. The left graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper right quadrant. The vertex is labeled “maximum”. The right graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. The vertex is labeled “minimum”.

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:


Continue With the Mobile App | Available on Google Play

[Attributions and Licenses]


This is a lesson from the tutorial, Introducing Quadratic Equations and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts