2.3 Finding Zeros of Linear and Quadratic Functions

To determine the x-intercepts of a polynomial function, we need to find the zeros of the function. A zero is a value of $x$ that satisfies $f(x)=0$. If $f(x)$ is a polynomial function, then a value of $x$ that satisfies $f(x)=0$ is called a zero or a root of the polynomial. Notice that since a zero makes the function equal to zero, a zero is the x-coordinate of an x-intercept. The maximum number of zeros of a polynomial is correlated with the degree of that polynomial. If the function is linear, it is a first-degree polynomial, and we expect there to be one zero maximum. For a second-degree polynomial, there will be two zeros maximum. For a third-degree polynomial, there will be three zeros maximum, and for an $n$th-degree polynomial, there will be a maximum of $n$ zeros. Note the word “maximum” is used here since it is possible that a polynomial could have fewer zeros than the degree of the polynomial.

Finding zeros is important for multiple reasons. One reason is that zeros help us graph functions. In addition, finding zeros will help us find the maximum and minimum of a function (we will talk about this in Chapter 5).

Zeros of Linear Functions

Let’s start with finding the zero of a linear function. Suppose that we have a function of the form

$f(x)=-3x+6$

To find a zero of this function, we need to set the equation equal to 0 and then solve for $x$.

$\begin{array}{rl}-3x+6& =0\\ -3x& =-6\\ x& =\frac{-6}{-3}=2\end{array}$

Since the zero of this function is $x=2$, we know that the x-intercept of this function is (2, 0).

Zeros of Quadratic Functions

How can we find the zeros of a quadratic function? We will discuss three methods: factoring, completing the square, and using the quadratic formula. Before using any of these methods to find the zeros, we must check if a quadratic equation has real solutions. The general form of a quadratic function is

$a{x}^2+bx+c=0$

Given an equation in this form, the number of real solutions (x-intercepts) for the quadratic equation is determined by

$b^2-4ac$

Suppose we have the following quadratic function:

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

How many real solutions does this function have? In order to determine this, we plug in the values of $a$, $b$, and $c$ to calculate $b^2-4ac$, as follows:

$\begin{array}{rl}{b}^2-4ac& =3^2-4(-2)(-5)\\ & =9-40\\ & =-31<0\end{array}$

Because $b^2-4ac<0$, there is no real solution.

Let’s look at another example. Suppose we have the following quadratic equation:

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

We can apply the same method to determine how many real solutions the equation has:

$\begin{array}{rl}{b}^2-4ac& =(-4{)}^2-4(1)(-2)\\ & =16+8\\ & =24>0\end{array}$

Because $b^2-4ac>0$, there are two real solutions.

Factoring

Factoring is usually the easiest method to find the zeros, although we cannot always factor quadratic functions easily. There are three things to consider first before factoring a quadratic equation.

1. If there is a common factor for all terms in the function, factor it out first.

2. If $a=1$ (the coefficient of $x^2$), use the product-and-sum method.

3. If $a\ne 1$, use the decomposition method.

Let’s use several examples to illustrate these three points.

First, let’s look at the following quadratic function:

$f(x)=x^2+2x$

Do $x^2$ and $2x$ have common factors? Yes, the common factor is $x$. So we factor $x$ to get

$f(x)=x(x+2)$

This function equals 0 when $x=0$ or $x=-2$. These are the zeros, the solutions to the quadratic function. So the x-intercepts of the function are (0, 0) and (-2, 0). Why is this true? As we know, anything multiplied by 0 is 0. If $x=0$, then $f(x=0)=0(0+2)=0$. Similarly, if $x=-2$ , then $f(x=0)-2(-2+2)=0$.

Now, look at the following quadratic function:

$f(x)=x^2-8x+12$

Do all the terms in the function have a common factor? The answer is no, so we move on to the next step to see if we can apply the product-and-sum method. What we need to find are two numbers such that the product of them is 12 and the sum of them is -8. This is because, if the factored function is $(x+a)(x+b)$, then:

$\begin{array}{rl}(x+a)(x+b)& =x^2+ax+bx+ab\\ & =x^2+(a+b)+ab\end{array}$

So, we want the sum of $a$ and $b$ to be -8 and the product of $a$ and $b$ to be 12. Here are all the possible combinations:

We know that the second-to-last combination is the right one to use to factor, so:

$\begin{array}{rl}f(x)& =x^2-8x+12\\ & =(x-2)(x-6)\end{array}$

So, the real solutions that make the function equal to zero are $x=2,6$.

The next quadratic function is a little bit different:

$f(x)=x^2-9$

While we could go through the steps to factor this equation, it is easier to use a simple rule of factoring for this one. Suppose that there is a general quadratic function without an $x$ term such that

$f(x)=x^2-c$

The function can be factored as

$f(x)=(x+\sqrt{c})(x-\sqrt{c})$

Applying this rule to the example above,

$\begin{array}{rl}f(x)& =x^2-9\\ & =(x+3)(x-3)\end{array}$

Next, suppose we have the following quadratic function:

$f(x)=-2x^2+8x+10$

First of all, we can factor the function by -2:

$f(x)=-2(x^2-4x-5)$

Now, we can apply the product-and-sum method to find two numbers for which the product is -5 and the sum is -4.

The second option is the correct one, so the factored quadratic function is as follows:

$f(x)=-2(x+1)(x-5)$

Therefore, the real solutions are $x=-1,5$.

Here is the last example to practice factoring. Consider the following quadratic function:

$f(x)=6x^2-4x-10$

First, we can factor the function by 2:

$f(x)=2(3x^2-2x-5)$

Now, since the coefficient of $x^2$ is not 1, we are going to apply the decomposition method. The idea is very similar to the product-and-sum method since some zeros may be fractions. In order to use this method, we first multiply the coefficient of $x^2$ and the constant, as follows:

$3\times (-5)=-15$

So what we want to find is a combination of two numbers for which the product is -15 and the sum is -2 (the coefficient of $x$). The combination of two numbers that we are looking for is -5 and 3. Next, we will decompose the $x$ term while keeping the $x^2$ term and the constant term as they are. How do we decompose the $x$ term? We are going to use the combination of two numbers, -5 and 3.

$-2x=-5x+3x$

Then, we are going to rewrite the quadratic function with the decomposition:

$\begin{array}{rl}f(x)& =2(3x^2-2x-5)\\ & =2(3x^2-5x+3x-5)\end{array}$

We can see something similar now, can’t we? We are going to factor $x$ out of $3x^2-5x$, and then factor $3x-5$ to complete the factoring:

$\begin{array}{rl}f(x)& =2(3x^2-5x+3x-5)\\ & =2\{x(3x-5)+1(3x-5)\}\\ & =2(x+1)(3x-5)\end{array}$

So we know that the real solutions of the quadratic function are $x=-1,\frac{5}{3}$ .

Completing the Square

Another method to find the real solutions of a quadratic function is by completing the square, just like we did to find the vertex of a quadratic function. Suppose we have the following quadratic function:

$f(x)=2x^2-6x-6$

Here is how we would complete the square for this function:

$\begin{array}{rl}f(x)& =2x^2-6x+14\\ & =2(x^2-6x)+14\\ & =2(x^2-6x+9-9)+14\\ & =2(x-3{)}^2-4\end{array}$

Next, we set the function equal to 0 and solve for $x$:

$\begin{array}{rl}2(x-3{)}^2-4& =0\\ 2(x-3{)}^2& =4\\ (x-3{)}^2& =2\\ x-3& =\pm \sqrt{2}\\ x& =3\pm \sqrt{2}\end{array}$

The answer, $x=3\pm \sqrt{2}$, is the real solution of the quadratic function.

The Quadratic Formula

The final method of finding the zeros is using the quadratic formula. The general form of a quadratic function is

$f(x)=a{x}^2+bx+c$

The real solutions of a quadratic function can be found by using the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

We can use the quadratic formula to solve any quadratic equation. If we cannot factor or complete the square, we can always use the quadratic formula. Notice the term in the square root, $b^2-4ac$. At the beginning of this section, we said that the term in the square root must be nonnegative in order for a quadratic function to have one or two real solutions. The reason for this is that if the term is negative, then the equation has imaginary solutions (with complex numbers), a subject that is beyond the scope of this course.

Let’s go over some examples. What is the solution of the following quadratic function?

$f(x)=-3x^2+x+2$

Simply plug the numbers into the quadratic formula to find the real solutions:

$\begin{array}{rl}x& =\frac{-1\pm \sqrt{1^2-4(-3)(2)}}{2(-3)}\\ & =\frac{-1\pm \sqrt{25}}{-6}\\ & =\frac{-1\pm 5}{-6}\\ & =-\frac{2}{3},1\end{array}$

Here is another example. Let’s find the real solutions of the following function:

$f(x)=x^2+x-5$

Again, plug the numbers into the quadratic formula to find the real solutions:

$\begin{array}{rl}x& =\frac{-1\pm \sqrt{1^2-4(1)(-5)}}{2(1)}\\ & =\frac{-1\pm \sqrt{21}}{2}\end{array}$

Graphing Quadratic Functions

Now that we know how to find the zeros, we can use them to graph quadratic functions. As we mentioned earlier, the zeros of quadratic functions are the x-coordinates of x-intercepts. So if we can find the zeros of a quadratic function and put the function in the form $f(x)=a(x-h{)}^2+k$, then we have all the information we need to graph the function.

Suppose that there is a quadratic function such that

$f(x)=x^2-3x-18$

First, we can factor the equation:

$\begin{array}{rl}f(x)& =x^2-3x-18\\ & =(x-6)(x+3)\end{array}$

So the x-intercepts are (-3, 0) and (6, 0). Now, let’s find the vertex of the function:

$\begin{array}{rl}f(x)& =x^2-3x-18\\ & ={(x-\frac{3}{2})}^2-\frac{81}{4}\end{array}$

The vertex of the parabola is $(\frac{3}{2}$ , $-\frac{81}{4})$. Since the coefficient of $x^2$ is positive, we know that the graph opens upward. Finally, the y-intercept is (0, -18). Given these points, we can graph this quadratic function as follows: