2.2 Quadratic Functions and Graphs

A quadratic function is a second-degree polynomial function. The general form of a quadratic function is

$y = a x^{2} + b x + c$

When we graph quadratic functions, we could plot the points one by one. However, it is better to complete the square, which will produce the following form, called the vertex form:

$y = a (x - h)^{2} + k$

This form will help us graph quadratic functions because it tells us important elements of the graph, which we will discuss later. We will first learn how to change the standard form of a quadratic function into vertex form.

Complete the Square and Vertex Form

Let’s start with learning how to complete the square of a quadratic function. Completing the square is the process of making a quadratic equation into a perfect square. That is, we want to rewrite $a x^{2} + b x + c$ in the form $a (x - h)^{2}$ . In order to do so, we want to make the standard form function into

$\begin{aligned} a x^{2} + b x + c = a (x^{2} + 2 h x + h^{2}) \\ = a (x + h)^{2} \end{aligned}$

Here are the steps to complete the square and transform the standard form equation into vertex form:

Factor both $x^{2}$ and $x$ by the coefficient of $x^{2}$ .
Divide the coefficient of $x$ by 2.
Square the outcome of step 2.
Add “zero.”
Complete the square.

We will use the following example to illustrate the process of completing the square and transforming the equation into vertex form. Suppose we have a function such that

$f (x) = x^{2} - 4 x$

where $f (x)$ means a function of $x$ .

To complete the square, we first need to factor by the coefficient of $x^{2}$ . In this example, the coefficient is 1. We can then move to the next step.

The second step is to divide the coefficient of $x$ by 2, which is -2.

The third step is to square the outcome of the second step. The square of -2 is 4.

In order to complete the square for $x^{2} - 4 x$ , we need +4. However, to keep the equation the same, if we add +4, then we have to subtract -4 on the same side of the equation, which is the fourth step:

$x^{2} - 4 x = x^{2} - 4 x + 4 - 4$

We can use parentheses to indicate the portion of the equation that can complete the square, and then any additional terms remain outside the parentheses as follows:

$\begin{aligned} f (x) & = x^{2} - 4 x \\ = (x^{2} - 4 x + 4) - 4 \end{aligned}$

Then we can complete the square within the parentheses:

$f (x) = x^{2} - 4 x$
$= (x^{2} - 4 x + 4) - 4$
$= (x - 2)^{2} - 4$

Notice that, in the equation on the second line, we added 4 to the function since we needed it to square the quadratic function, and then we subtracted 4 to cancel out the addition (this is equivalent to adding 0 to the function). The last line is the vertex form of the quadratic equation. Focusing on the part within the parentheses, in the third line, $x$ is added (adding -2 is equivalent to subtracting 2) to half of the coefficient of $x$ from the second line, and then the result is squared.

Let’s look at one more example. Suppose that we have the following quadratic function:

$f (x) = 2 x^{2} + 8 x - 3$

Since the coefficient of $x^{2}$ is 2, we will factor both the $x^{2}$ and $x$ terms by 2 to get

$f (x) = 2 (x^{2} + 4 x) - 3$

Next, since the coefficient of $x$ is 4, we divide it by 2 to get 2. Then, we square it to get 4. This means that we need a 4 to complete the square. So we will add “zero” to the equation by adding 4 and then subtracting 4 within the parentheses:

$f (x) = 2 (x^{2} + 4 x + 4 - 4) - 3$

We want only $x^{2} + 4 x + 4$ in the parentheses, so we use the distributive property to multiply -4 by 2 and then add the result to the -3 ouside the parentheses. We can now complete the square:

$\begin{aligned} f (x) & = 2 (x^{2} + 4 x + 4 - 4) - 3 \\ = 2 (x^{2} + 4 x + 4 - 4) - 3 - 8 \\ = 2 (x^{2} + 4 x + 4) - 11 \\ = 2 (x + 2)^{2} - 11 \end{aligned}$

Let’s try one more example to make sure we can complete the square. Suppose we have a quadratic function such that

$f (x) = - 3 x^{2} + 10 x - 5$

Again, we will use the same steps:

Factor both $x^{2}$ and $x$ by the coefficient of $x^{2}$ .
Divide the coefficient of $x$ by 2.
Square the outcome of step 2.
Add “zero.”
Complete the square.

$\begin{aligned} f (x) & = - 3 x^{2} + 10 x - 5 \\ = - 3 (x^{2} - \frac{10}{3} x) - 5 \\ = - 3 (x^{2} - \frac{10}{3} x + \frac{25}{9} - \frac{25}{9}) - 5 \\ = - 3 (x^{2} - \frac{10}{3} x + \frac{25}{9}) - 5 + \frac{25}{3} \\ = - 3 (x - \frac{5}{3}) + \frac{10}{3} \end{aligned}$

Quadratic Function Graph

It is important to remember that the graph of any quadratic function is a transformation of the graph of the function $f (x) = x^{2}$ . Even though it may look complicated at first, let’s look at the general form of the quadratic function:

$\begin{aligned} f (x) & = a x^{2} + b x + c \\ = a (x^{2} + \frac{b}{a} x) + c \\ = a (x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}} - \frac{b^{2}}{4 a^{2}}) + c \\ = a (x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}}) + c - \frac{b^{2}}{4 a} \\ = a {(x + \frac{b}{2 a})}^{2} + \frac{4 a c - b^{2}}{4 a} \\ = a (x - h)^{2} + k \end{aligned}$

You may wonder why it is important to transform the equation to the vertex form. The vertex form of a quadratic function tells us some of the values we need to graph the function. For example, if $f (x) = a (x - h)^{2} + k$ , then:

The coefficient $a$ tells us which way the graph opens: If $a > 0$ , the graph opens upward; if $a < 0$ , the graph opens downward.
The constants $h$ and $k$ tell us the vertex of the graph. The vertex of the graph is ( $h$ , $k$ ).
The vertical line that goes through $h$ is the parabola’s axis of symmetry, $x = h$ .

Recall what you have learned in previous math classes about graphing a quadratic function. The graph of a quadratic function is the shape of a parabola.

Pay attention to the sign of the value $h$ . If $h$ is negative, as shown below, then the vertex is ( $h$ , $k$ ).

$f (x) = a (x - h)^{2} + k$

If $h$ is positive, as shown below, then the vertex is ( $- h$ , $- k$ ).

$f (x) = a (x + h)^{2} - k$

Thus, the sign of x-coordinate of the vertex is the opposite of the sign of the value in the function, while the sign of y-coordinate of the vertex has same sign as the value in the function.

Figure 2.1: Graphs of Quadratic Functions

How can we find the x-intercepts and the y-intercept of a quadratic equation? To find the y-intercept, we plug 0 in for $x$ in the function and solve for $y$ . The result is the y-coordinate of the intercept (the x-coordinate of the intercept is 0). In order to find the x-intercepts, we need to learn how to find zeros. We will discuss finding zeros in next section.

Examples

Let’s go through some examples to master graphing quadratic functions.

Example 1

Graph the quadratic function $f (x) = 3 x^{2} - 9 x - 4$ . On the graph, specify the vertex and y-intercept.

First, we need to square the quadratic function.

$\begin{aligned} f (x) & = 3 x^{2} - 9 x - 4 \\ = 3 (x^{2} - 3 x) - 4 \\ = 3 (x^{2} - 3 x + \frac{9}{4} - \frac{9}{4}) - 4 \\ = 3 (x^{2} - 3 x + \frac{9}{4}) - 4 - \frac{27}{4} \\ = 3 {(x - \frac{3}{2})}^{2} - \frac{43}{4} \end{aligned}$

The y-coordinate of the y-intercept is found as follows:

$f (x = 0) = 3 (0)^{2} - 9 (0) - 4 = - 4$

The graph of this function is shown below:

Figure 2.2: Graph of

f (x) = 3 (x - \frac{3}{2}) - \frac{43}{4}

Example 2

Graph the quadratic function $f (x) = - x^{2} - 6 x + 2$ . On the graph, specify the vertex and y-intercept.

$\begin{aligned} f (x) & = - x^{2} - 6 x + 2 \\ = - (x^{2} + 6 x) + 2 \\ = - (x^{2} + 6 x + 9 - 9) + 2 \\ = - (x^{2} + 6 x + 9) + 2 + 9 \\ = - (x + 3)^{2} + 11 \end{aligned}$

The y-coordinate of the y-intercept is found as follows:

$f (x = 0) = - 0^{2} - 6 (0) + 2 = 2$

The graph of this function is shown below:

Figure 2.3: Graph of

f (x) = - (x + 3) + 11

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