# Minimum Values: Definition & Concept

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• 0:00 Exercise
• 0:18 How to Determine Minimum Value
• 2:36 Real World Examples
• 4:01 Lesson Summary

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Lesson Transcript
Instructor: Jennifer Beddoe
The minimum value of a quadratic function is the low point at which the function graph has its vertex. This lesson will define minimum values and give some example problems for finding those values. A quiz will complete the lesson.

## Exercise

The minimum value of a function is the place where the graph has a vertex at its lowest point. In the real world, you can use the minimum value of a quadratic function to determine minimum cost or area. It has practical uses in science, architecture and business.

## How to Determine Minimum Value

There are three methods for determining the minimum value of a quadratic equation. Each of them can be useful in determining the minimum.

The first way is by using a graph. You can find the minimum value visually by graphing the equation and finding the minimum point on the graph. The y-value of the vertex of the graph will be the minimum. This is especially easy when you have a graphing calculator that can do most of the work for you. Looking at this graph, you can see that the minimum point of the graph is at y = -3.

The second way to find the minimum value comes when you have the equation y = ax^2 + bx + c. If your equation is in the form y = ax^2 + bx + c, you can find the minimum by using the equation min = c - b^2/4a.

The first step is to determine whether your equation gives a maximum or minimum. This can be done by looking at the x^2 term. If this term is positive, the vertex point will be a minimum; if it is negative, the vertex will be a maximum. After determining that you actually will have a minimum point, use the equation to find it.

Let's do an example. Find the minimum point of 3x^2 + 12x + 2.

Since the term with the x^2, or 'a' term, is positive, you know there will be a minimum point. To find it, plug the values into the equation min = c - b^2/4a.

That gives us min = 2 - 12^2/4(3)

This simplifies to min = 2 - 144 / 12, which can be further simplified to min = 2 - (12), or min = -10.

The third way to find the minimum value is using the equation y = a(x - h)^2 + k.

As with the last equation, the a term in this equation must be positive for there to be a minimum. If the a term is positive, the minimum can be found at k. No equation or calculation is necessary; the answer is just k.

Let's look at an example and find the minimum of the equation (x + 13)^2 + 2.

Since the a term is positive, there will be a minimum at y = 2.

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