Quadratic Functions: Maximum and Minimum Values

Published: Dec 25th, 2020

Learning Objectives
By the end of this section, you should be able to:
1. Find the vertex of a function using the "completing the square" method
2. Determine whether a point is a maximum/minimum

In this lesson, we're going to the E X T R E M E S

Us, as we go to the extremes.

The extreme values of a quadratic function, that is!

In Lesson #13, we entered the lemonade business and maximized our profits. 

You may remember we introduced two methods to calculate the value at the peak of a function:

  1. Finding its roots and taking the average of the roots
  2. Using the expression \(\frac{-b}{2a}\)

In this case, we're going to introduce the logic behind these methods and a third way to find the vertex.

As you may recall from Lesson #4, quadratic functions are parabola-shaped. We also discussed that these functions can be expressed in vertex form\(a(x-p)^2+q\), where \((p,q)\) are the coordinates of the vertex, the extreme value of the function (either its maximum or minimum value, depending on if opens downward or upward, respectively). 

Now, what if we want to convert a function from standard form \(ax^2+bx+c\) into vertex form? Great question, I'll show you!

Convert \(x^2+4x+9\) to vertex form.

Before we do this, let's dissect the vertex form, shall we?

Consider the function \((x-1)^2+4\). Let's expand it:

\(=x^2 - 2x + \color{red}{1}+4\)

\(=x^2 - 2x + 5\)

What we want to do, is bring this function back to \((x-1)^2+4\)

Why the 1 though? Where did the one come from? 

Well, when we convert a function to vertex form, we need the first term \((x-p)^2\), to contain the largest perfect square possible.

How do we find the largest perfect square possible? Well, it involves:

\((\frac{b}{2})^2\)

We want to take our \(b\), value, -2 in this case, and substitute into the expression above:

\((\frac{-2}{2})^2\)

\(=1\)

There's the 1 again!

Now, how we convert this back into vertex form? Let's add the one into our expression

\(=x^2 - 2x \color{red}{+1}+ 5\) Not so fast! We can't just add things to our equations, as otherwise they won't represent the original equation anymore! We need to balance things out:

\(=x^2 - 2x \color{red}{+1}+ 5 \color{blue}{-1}\) Aha, that's much better

Now, factor the first three terms:

\(=(x-1)^2+ 5 \color{blue}{-1}\)

Looks good so far. Now simplify:

\(=(x-1)^2+ 4\)

With that, we're back to vertex form. The conversion from standard \(ax^2+bx+c\) to vertex \(a(x-p)^2+q\) form is called completing the square.

Now, back to our example:\(x^2+4x+9\)

What's our perfect square value? Let's find out:

\((\frac{4}{2})^2\)

\(=4\)

Looks like it's four! Add and remove it as we did before:

\(x^2+4x\color{red}{+4}+9\color{blue}{-4}\)

Factor:

\((x+2)^2+9\color{blue}{-4}\)

Simplify:

\((x+2)^2+5\)

And that's our vertex form! If you want, you can try to plot both on a graph. You will find that they're the same function.

Now, let's extend this idea a little bit:

Find the maximum/minimum values of \(x^2+4x+9\)

So, at this point we have a value for our vertex. This is an extreme value for the function either a maximum or a minimum.

But which is it? A maximum or a minimum?

Well, you might notice that the function is written in standard form. In this case \(a>0\), so the function opens upwards (smiley face). Therefore, this point is a minimum of the function, since the function will increase in the negative and positive directions relative to this point!

Now, we know that our minimum value is (-2,5), but what is our maximum? Well, that depends on the domain.

In this case, our domain is not restricted, so there is no maximum value; we'll call it \(+\infty\) (positive infinity). This is because, in either direction, the function will continue to increase. Since we haven't limited our domain, the function spans all real numbers, and will continue to increase infinitely!

What if we restricted our domain to [-4,1] (note that the square brackets imply an inclusive domain, meaning that we're including -4 and 0 in the domain)?

Well, then we need to find the value at either endpoint, and compare to see which is larger!

\(f(-4)\) = \((-4)^2+4(-4)+9\)

\(= 16 - 16 + 9\)

\(=9\)

\(f(1) = (1)^2+4(1) +9\)

\(= 1 + 4 + 9\)

\(=14\)

Since 14 is greater than 9, then within this domain, (1,14) is our absolute maximum value. 

More on maxima and minima will be discussed in upcoming lessons, but hopefully this has been a good introduction. Be sure to try some practice problems!

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