Graphing Quadratic Equations Using Factoring

A quadratic equation can be defined as a polynomial equation of degree 2. The standard form of a quadratic equation is:

$0={ax}^2+bx+c$

In this form, a, b, and c must all be real numbers and $a\ne 0$ . If we replace the 0 with a y, we get a quadratic function:

$y={ax}^2+bx+c$

The graph of such a function will always be a parabola. These graphs can be tricky to sketch, but factoring the quadratic gives us all of the information we need to do so successfully. In this article. we'll look at graphing quadratic equations using factoring. Let's get started!

The procedure for graphing quadratic equations using factoring

The points where any parabola intersects the x-axis will be the solutions to the equation ${ax}^2+bx+c=0$ . Put another way, if we can factor any quadratic equation in $0={ax}^2+bx+c$ format into $0=(x-p)(x-q)$ , we know by the zero product property that if $(x-p)(x-q)=0$ , either ∈. This means that p and q are the solutions to the equation ${ax}^2+bx+c$ and the x-intercepts when we graph the quadratic equation. It's an easy way to get two specific, important points to include in our graph.

We're not done yet, though. Assuming a quadratic equation is written in $0={ax}^2+bx+c$ form, c tells us what the y-intercept will be. More specifically, the constant term places the y-intercept at $\left(0,c\right)$ , giving us a third specific point to work with. Likewise, the value of a tells us whether our parabola will open up or down. If a is positive, the parabola opens upward. If it's negative, the parabola opens downward.

We can also figure out the vertex of the parabola by factoring the quadratic. The x-coordinate of the vertex of a parabola will always be the midpoint between the two x-intercepts, allowing us to use the following formula to determine it:

$\frac{p+q}{2}$

Once we have the x-coordinate, we can determine the y-coordinate by plugging the x value into the original equation and solving for y. With four specific points including both x-intercepts and the vertex of the parabola, we can create an accurate sketch of the graph quickly and easily.

Practical examples of graphing quadratic equations using factoring

All of the information above may have been difficult to follow in general terms, so let's try graphing a couple of quadratic equations this way to see how it works. Let's begin with: $y=x^2-8x+12$

As you may recall, the process for factoring a quadratic equation requires finding a factor pair with a product of c and a sum of b when the equation is in $0={ax}^2+bx+c$ format. In this equation, our b value is -8 and our c value is 12, which means that we need to find two numbers with a sum of -8 and a product of 12. The easiest way to do this is often listing factor pairs with a product of 12 until we find a combination that also has a sum of -8:

$4\left(3\right)$

$-4\left(-3\right)$

$6\left(2\right)$

$-6\left(-2\right)$

There are other factor pairs with a product of 12, but we can stop here since -6 and -2 produce the product and sum we're looking for. Next, we can rewrite our quadratic equation:

$x^2-8x+12=\left(x-6\right)\left(x-2\right)$

By the zero product property, we now know that either $x-6=0$ or $x-2=0$ . Let's solve for x:

$x-6=0$

$x=6$

$x-2=0$

$x=2$

The roots of our equation are 6 and 2, which means that our x-intercepts will be $\left(6,0\right)$ and $\left(2,0\right)$ .

Next, we need to figure out the vertex of the parabola. We can find its x-coordinate using the formula:

$\frac{p+q}{2}$

$\frac{6+2}{2}$

$\frac{8}{2}$

$4$

Since 4 is the x-coordinate of the vertex, we can plug it into the original equation to find its y-value:

$y=x^2-8x+12$

$y=4^2-8\left(4\right)+12$

$y=-4$

The vertex of the parabola is $\left(4,-4\right)$ . We now have all of the information we need to sketch our graph:

Note that while we know the y-intercept will be $\left(0,12\right)$ , the scale of the graph above doesn't show us that point.

Practice questions on graphing quadratic equations using factoring

a. Graph the function $f\left(x\right)=-x^2-2x+8$ using factoring.

Factoring the quadratic equation is the first step when graphing a function using factoring, so let's compare our equation to the standard ${ax}^2+bx+c$ format. The value of a is negative, so the parabola will open downward. The c value is 8, which means the y-intercept will be $\left(0,8\right)$ .

First, let's factor out the -1:

$-x^2-2x+8=-1(x^2+2x-8)$

Next, we need to factor the equation in the parenthesis. We need a factor pair that adds to 2 and multiplies to -8. 4 and -2 work, giving us:

$x^2+2x-8=(x+4)(x-2)$

Using the Zero Product Property, we know that $x+4=0$ or $x-2=0$ . We can therefore solve both equations to get our x-intercepts:

$x+4=0$

$x=4$

$x-2=0$

$x=2$

Our x-intercepts are $\left(-4,0\right)$ and $\left(2,0\right)$ . Now, it's time to start working on the vertex of our parabola. The formula is:

$\frac{p+q}{2}$

$\frac{-4+2}{2}$

$-1$

Next, we plug the x-coordinate of -1 into the original equation to determine the vertex's y-coordinate:

$y=-x^2-2x+8$

$y=-{\left(-1\right)}^2-2(-1)+8$

$y=9$

The coordinates of the vertex of the parabola are $\left(-1,9\right)$ . Combining this with the y-intercept of $\left(0,8\right)$ and the two x-intercepts of $\left(-4,0\right)$ and $\left(2,0\right)$ , we have four points with which to plot our graph. Connect them with a smooth, continuous line and it should look something like this:

Topics related to the Graphing Quadratic Equations Using Factoring

Quadratic Equations, Solving by Factoring

Difference of Squares

Advanced Factoring

Flashcards covering the Graphing Quadratic Equations Using Factoring

Algebra 1 Flashcards

College Algebra Flashcards

Practice tests covering the Graphing Quadratic Equations Using Factoring

Algebra 1 Diagnostic Tests

College Algebra Diagnostic Tests

Graphing quadratic equations using factoring is simpler with Varsity Tutors

Using factoring to graph quadratic equations is easier than other methods available to us, but it still entails multiple steps that can throw students off. If the student in your life needs help memorizing the formula for the x-coordinate of the vertex of a parabola or struggles to find an appropriate factor pair, a private tutor can focus on their specific areas of opportunity to optimize their study efficiency. Please reach out to the Educational Directors at Varsity Tutors for more information on the benefits of tutoring and how easy it can be to get started.