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Minimum Values: Definition & Concept

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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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