Curved antennas, such as the ones shown in the photo, are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.

An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr)

The simplest example of a quadratic function, that you have likely come across before, is [latex]f\left(x\right)={x}^{2}[/latex]. Before we talk about more general equation of a quadratic function, we will look at its graph.

Characteristics of Parabolas

The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry.

The [latex]y[/latex]-intercept is the point at which the parabola crosses the [latex]y[/latex]-axis. The [latex]x[/latex]-intercepts are the points at which the parabola crosses the [latex]x[/latex]-axis. If they exist, the [latex]x[/latex]-intercepts represent the zeros, or roots, of the quadratic function, the values of [latex]x[/latex] at which [latex]y=0[/latex].

Example: Identifying the Characteristics of a Parabola

Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown below.

Show Solution

The vertex is the turning point of the graph. We can see that the vertex is at [latex](3,1)[/latex]. The axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is [latex]x=3[/latex]. This parabola does not cross the [latex]x[/latex]-axis, so it has no zeros. It crosses the [latex]y[/latex]-axis at (0, 7) so this is the [latex]y[/latex]-intercept.

Equations of Quadratic Functions

The general form of a quadratic function presents the function in the form

[latex]f\left(x\right)=a{x}^{2}+bx+c[/latex]

where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are real numbers and [latex]a\ne 0[/latex]. If [latex]a>0[/latex], the parabola opens upward. If [latex]a<0[/latex], the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.

The axis of symmetry is defined by [latex]x=-\dfrac{b}{2a}[/latex]. If we use the quadratic formula, [latex]x=\dfrac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex], to solve [latex]a{x}^{2}+bx+c=0[/latex] for the [latex]x[/latex]-intercepts, or zeros, we find the value of [latex]x[/latex] halfway between them is always [latex]x=-\dfrac{b}{2a}[/latex], the equation for the axis of symmetry.

The figure below shows the graph of the quadratic function written in general form as [latex]y={x}^{2}+4x+3[/latex]. In this form, [latex]a=1,\text{ }b=4[/latex], and [latex]c=3[/latex]. Because [latex]a>0[/latex], the parabola opens upward. The axis of symmetry is [latex]x=-\dfrac{4}{2\left(1\right)}=-2[/latex]. This also makes sense because we can see from the graph that the vertical line [latex]x=-2[/latex] divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, [latex]\left(-2,-1\right)[/latex]. The [latex]x[/latex]-intercepts, those points where the parabola crosses the [latex]x[/latex]-axis, occur at [latex]\left(-3,0\right)[/latex] and [latex]\left(-1,0\right)[/latex].

The standard form of a quadratic function presents the function in the form

[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]

where [latex]\left(h,\text{ }k\right)[/latex] is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.

Figure 5

As with the general form, if [latex]a>0[/latex], the parabola opens upward and the vertex is a minimum. If [latex]a<0[/latex], the parabola opens downward, and the vertex is a maximum. Figure 5 is the graph of the quadratic function written in standard form as [latex]y=-3{\left(x+2\right)}^{2}+4[/latex]. Since [latex]x-h=x+2[/latex] in this example, [latex]h=-2[/latex]. In this form, [latex]a=-3,\text{ }h=-2[/latex], and [latex]k=4[/latex]. Because [latex]a<0[/latex], the parabola opens downward. The vertex is at [latex]\left(-2,\text{ 4}\right)[/latex].

The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.

Given a quadratic function in general form, find the vertex of the parabola.

One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, [latex]k[/latex], and where it occurs, [latex]h[/latex]. If we are given the general form of a quadratic function:

[latex]f(x)=ax^2+bx+c[/latex]

We can define the vertex, [latex](h,k)[/latex], by doing the following:

• Identify [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex].
• Find [latex]h[/latex], the [latex]x[/latex]-coordinate of the vertex, by substituting [latex]a[/latex] and [latex]b[/latex] into [latex]h=-\dfrac{b}{2a}[/latex].
• Find [latex]k[/latex], the [latex]y[/latex]-coordinate of the vertex, by evaluating [latex]k=f\left(h\right)=f\left(-\dfrac{b}{2a}\right)[/latex]

Finding the Domain and Range of a Quadratic Function

Any number can be the input value of a quadratic function. Therefore the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum at the vertex, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all [latex]y[/latex]-values greater than or equal to the [latex]y[/latex]-coordinate of the vertex or less than or equal to the [latex]y[/latex]-coordinate at the turning point, depending on whether the parabola opens up or down.

Intercepts of Quadratic Functions

When graphing parabolas, it’s often helpful to find intercepts of quadratic function. Recall that we find the [latex]y[/latex]-intercept of a function by evaluating the function at an input of zero, and we find the [latex]x[/latex]-intercepts at locations where the output is zero. Notice that the number of [latex]x[/latex]-intercepts can vary depending upon the location of the graph.

Number of [latex]x[/latex]-intercepts of a parabola

Mathematicians also define [latex]x[/latex]-intercepts as roots of the quadratic function. This is where you will apply the methods of solving quadratic equations you reviewed in the Quadratic equations section.

Example: Finding the y– and x-Intercepts of a Parabola

Find the [latex]y[/latex]– and [latex]x[/latex]-intercepts of the quadratic [latex]f\left(x\right)=3{x}^{2}+5x - 2[/latex].

Show Solution

We find the [latex]y[/latex]-intercept by evaluating [latex]f\left(0\right)[/latex].

[latex]f\left(0\right)=3{\left(0\right)}^{2}+5\left(0\right)-2=-2[/latex]

So the [latex]y[/latex]-intercept is at [latex]\left(0,-2\right)[/latex].

For the [latex]x[/latex]-intercepts, or roots, we find all solutions of [latex]f\left(x\right)=0[/latex].

[latex]3{x}^{2}+5x - 2=0[/latex]

In this case, the quadratic can be factored easily, providing the simplest method for solution.

[latex]\left(3x - 1\right)\left(x+2\right)=0[/latex]
[latex]\begin{align}&3x - 1=0&x+2=0\end{align}[/latex]
[latex]x=\frac{1}{3}\hspace{8mm}[/latex] or [latex]\hspace{8mm}x=-2[/latex]

So the roots are at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].

Analysis of the Solution

By graphing the function, we can confirm that the graph crosses the [latex]y[/latex]-axis at [latex]\left(0,-2\right)[/latex]. We can also confirm that the graph crosses the [latex]x[/latex]-axis at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].

In the above example the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.

Example: Finding the Roots of a Parabola

Find the [latex]x[/latex]-intercepts of the quadratic function [latex]f\left(x\right)=2{x}^{2}+4x - 4[/latex].

Show Solution

We begin by solving for when the output will be zero.

[latex]0=2{x}^{2}+4x - 4[/latex]

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]

We know that [latex]a=2[/latex]. Then we solve for [latex]h[/latex] and [latex]k[/latex].

[latex]\begin{align}&h=-\dfrac{b}{2a}=-\dfrac{4}{2\left(2\right)}=-1\\[2mm]&\end{align}[/latex]

[latex]\begin{align}k&=f\left(h\right)\\&=f\left(-1\right)\\&=2{\left(-1\right)}^{2}+4\left(-1\right)-4\\&=-6\end{align}[/latex]

So now we can rewrite in standard form.

[latex]f\left(x\right)=2{\left(x+1\right)}^{2}-6[/latex]

We can now solve for when the output will be zero.

[latex]\begin{align}&0=2{\left(x+1\right)}^{2}-6 \\ &6=2{\left(x+1\right)}^{2} \\ &3={\left(x+1\right)}^{2} \\ &x+1=\pm \sqrt{3} \\ &x=-1\pm \sqrt{3} \end{align}[/latex]

The graph has [latex]x[/latex]–intercepts at [latex]\left(-1-\sqrt{3},0\right)[/latex] and [latex]\left(-1+\sqrt{3},0\right)[/latex].

Analysis of the Solution

We can check our work by graphing the given function on a graphing utility and observing the roots.