Graphing Quadratic Equations Using Tables

The simplest form of a quadratic equation is [latex]y=ax^2[/latex]. This is also referred to as the parent equation of any quadratic equation. The basic shape of a quadratic equation is a parabola. It has a vertex where the parabola turns and a line of symmetry that runs through the parabola and splits the graph into two mirror images. We discovered all of this in the last section by determining solutions of the equation and plotting the solution points. We can use this technique for any quadratic equation.

Example

Complete a table of values for the equation [latex]y=2x^2[/latex], then graph the equation.

Solution

To create a table of values, we can choose any [latex]x[/latex]-values and find the corresponding [latex]y[/latex]-values.

[latex]x[/latex]	[latex]y=2x^2[/latex]
[latex]0[/latex]	[latex]y=2(0)^2=0[/latex]
[latex]1[/latex]	[latex]y=2(1)^2=2[/latex]
[latex]-1[/latex]	[latex]y=2(-1)^2=2[/latex]
[latex]2[/latex]	[latex]y=2(2)^2=8[/latex]
[latex]-2[/latex]	[latex]y=2(-2)^2=8[/latex]

To create the graph, we plot the points and join the dots.

Notice that it is still a parabola. All quadratic equations take the shape of a parabola.

Example

Complete a table of values for the equation [latex]y=-x^2+4x-3[/latex], then graph the equation. State the vertex and axis of symmetry.

Solution

To create a table of values, we can choose any [latex]x[/latex]-values and find the corresponding [latex]y[/latex]-values.

[latex]x[/latex]	[latex]y=-x^2+4x-3[/latex]
[latex]0[/latex]	[latex]y=(0)^2+4(0)-3=-3[/latex]
[latex]1[/latex]	[latex]y=-(1)^2+4(-1)-3=0[/latex]
[latex]-1[/latex]	[latex]y=-(-1)^2+4(-1)-3=-8[/latex]
[latex]2[/latex]	[latex]y=-(2)^2+4(2)-3=1[/latex]
[latex]-2[/latex]	[latex]y=-(-2)^2+4(-2)-3=-15[/latex]

To create the graph, we plot the points.

Unfortunately, the points we chose do not show any symmetry or a turning point. However, the graph looks like it will turn to the right of [latex]x=2[/latex], so let’s find a few other points that lie to the right of [latex]x=2[/latex].

[latex]x[/latex]	[latex]y=-x^2+4x-3[/latex]
[latex]3[/latex]	[latex]y=-(3)^2+4(3)-3=0[/latex]
[latex]4[/latex]	[latex]y=-(4)^2+4(4)-3=-3[/latex]
[latex]5[/latex]	[latex]y=-(5)^2+4(5)-3=-8[/latex]

Now we have some symmetry and can join the dots to create the graph.

 

The vertex is at [latex](2, 1)[/latex] and the axis of symmetry is the vertical line [latex]x=2[/latex].

Notice also, that the point [latex](-2, -15)[/latex] has a twin as a mirror image at [latex](6, -15)[/latex].

In the first example, the parabola opens upwards, while in the second example, the parabola opens downwards. This is determined by the value of [latex]a[/latex] in the equation [latex]y=ax^2+bx+c[/latex]. When [latex]a>0[/latex], the parabola opens upwards. When [latex]a<0[/latex], the parabola opens downwards.  Notice that [latex]a\neq0[/latex] because that would turn the quadratic equation [latex]y=ax^2+bx+c[/latex] into a linear equation [latex]y=bx+c[/latex].

Try It

Complete a table of values for the equation [latex]y=-x^2+7[/latex], then graph the equation. State the vertex and axis of symmetry.

Show Answer

First notice that [latex]a=-1[/latex] so the parabola will open downwards.

To complete a table of values we can choose any [latex]x[/latex]-values:

[latex]x[/latex]	[latex]y=-x^2+7[/latex]
-3	[latex]y=-(-3)^2+7=-2[/latex]
-2	[latex]y=-(-2)^2+7=3[/latex]
-1	[latex]y=-(-1)^2+7=6[/latex]
0	[latex]y=-(0)^2+7=7[/latex]
1	[latex]y=-(1)^2+7=6[/latex]
2	[latex]y=-(2)^2+7=3[/latex]
3	[latex]y=-(3)^2+7=-2[/latex]

From the graph the vertex is at [latex](0, 7)[/latex] and the line of symmetry is the vertical line [latex]x=0[/latex].

Sometimes it can be messy to draw a graph using just solutions. Deciding which [latex]x[/latex]-values to choose can be bothersome. It can also be impossible to determine exactly where the axis of symmetry and vertex lie if they do not contain integer values. That’s why we use other features of the graph to help us.  For example, it would be helpful to know exactly where the vertex and axis of symmetry lie. Or, where the graph crosses the axes. So, let’s discover how to determine such features.

Features of Parabolas

Intercepts: The [latex]y[/latex]-Intercept

The [latex]y[/latex]-intercept of any graph is found by setting [latex]x=0[/latex] in the equation of the graph and solving for [latex]y[/latex]. For a parabola with equation [latex]y=ax^2+bx+c[/latex], setting [latex]x=0[/latex] results in [latex]y=c[/latex]. Consequently, the [latex]y[/latex]-intercept of any parabola is always the point [latex](0, c)[/latex].

Intercepts: The [latex]x[/latex]-Intercepts

The [latex]x[/latex]-intercepts of a parabola occur where the parabola crosses the [latex]x[/latex]-axis (figure 1). Specifically, this is where [latex]y=0[/latex]. So to find the [latex]x[/latex]-intercepts of a parabola, we must solve the equation [latex]ax^2+bx+c=0[/latex].

Figure 1. [latex]x[/latex]-intercepts

Provided this quadratic equation factors, we can solve it using the zero-factor property.  If the equation does not factor, there are alternative ways of solving it that will be taught in the next course.

Parabolas whose equations factor have [latex]x[/latex]-intercepts. But not all parabolas have [latex]x[/latex]-intercepts. Consider the parabolas in figure 2:

Figure 2. Parabolas with 2, 1, or 0 [latex]x[/latex]-intercepts.

Parabolas can have two [latex]x[/latex]-intercepts, one [latex]x[/latex]-intercept, or no [latex]x[/latex]-intercepts.

If a parabola does not intersect with the [latex]x[/latex]-axis,and therefore has no [latex]x[/latex]-intercepts, there are complex number solutions to the equation [latex]y=ax^2+bx+c=0[/latex]. Such solutions cannot be graphed on the real number line, so will not appear on the graph.

Axis of Symmetry and the Vertex

The axis of symmetry and the vertex are very important features of a parabola so being able to find them will be extremely useful for graphing.

For example, consider the equation [latex]y=2x^2-3x+4[/latex]. To find the axis of symmetry calculate [latex]x=-\frac{b}{2a}=-\frac{-3}{2\cdot 2}=\frac{3}{4}[/latex]. The axis of symmetry is the vertical line with equation [latex]x=\frac{3}{4}[/latex].

The axis of symmetry passes through the vertex, so the [latex]x[/latex]-coordinate of the vertex is also [latex]\frac{3}{4}[/latex]. To find the [latex]y[/latex]-coordinate, we substitute [latex]x=\frac{3}{4}[/latex] into the original equation [latex]y=2x^2-3x+4[/latex]:

[latex]\begin{equation}\begin{aligned}y&=2x^2-3x+4 \\ y&=2 {\left (\frac{3}{4}\right )}^{2}-3\cdot\frac{3}{4}+4 \\ y&=2\cdot\frac{9}{16}-\frac{9}{4}+4\\y&=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\y&=\frac{23}{8}\end{aligned}\end{equation}[/latex]

The vertex is at the point [latex]\left(\dfrac{3}{4},\dfrac{23}{8}\right)[/latex]. This is verified by the graph in figure 3:

Figure 3. Graph of [latex]y=2x^2-3x+4[/latex]

Putting it all Together

We have learned some important things about the graphs of quadratic equations that will make it easier for us to create a graph.  These features of a parabola are summarized below:

We can use the properties of parabolas to help us graph a quadratic equation of the form [latex]y=ax^2+bx+c[/latex] without having to calculate an exhaustive table of values.

Example

Graph [latex]y=−2x^{2}+3x–3[/latex].

Solution

Let’s start by considering the features of a parabola.

Opening:

[latex]a=-2[/latex] so the parabola opens downwards.

Since [latex]|a|>1[/latex] the graph will be narrower than the graph of [latex]y=x^2[/latex].

[latex]y[/latex]-intercept:

Since [latex]c=-3[/latex], the [latex]y[/latex]-intercept will be [latex](0, -3)[/latex].

Axis of symmetry:

[latex]x=\frac{-b}{2a}=\frac{-3}{2\cdot (-2)}=\frac{3}{4}[/latex].

Vertex:

[latex]x=\frac{3}{4}[/latex], so [latex]y=−2x^{2}+3x–3=-2{\left (\frac{3}{4}\right )}^{2}+3\cdot \frac{3}{4}-3=-2\cdot \frac{9}{16}+\frac{9}{4}-3=\frac{-9}{8}+\frac{9}{4}-3=\frac{-9+18-24}{8}=\frac{-15}{8}[/latex].  The vertex is the point [latex]\left (\frac{3}{4}, -\frac{15}{8}\right )[/latex].

[latex]x[/latex]-intercepts: 

[latex]−2x^{2}+3x–3[/latex] does not factor, so we cannot find the [latex]x[/latex]-intercepts.

Let’s graph everything we have:

Notice that the point [latex](0, -3)[/latex] has a mirror image on the other side of the axis of symmetry.

This twin has the same [latex]y[/latex]-value of [latex]-3[/latex]. To find the [latex]x[/latex]-value, move the same number of units past the axis of symmetry to the right.

The point [latex](0, -3)[/latex] is [latex]\frac{3}{4}[/latex] of a unit to the left of the axis, so its twin is [latex]\frac{3}{4}[/latex] of a unit to the right of the axis at [latex]\left (\frac{3}{2}, -3\right )[/latex].

From the information we have found so far, we have the turning point and direction of the graph. Now let’s find some solutions and use their twins to plot more points.

 

 

[latex]x[/latex]	[latex]y=−2x^{2}+3x–3[/latex]
[latex]−1[/latex]	[latex]−8[/latex]
[latex]0[/latex]	[latex]−3[/latex]
[latex]1[/latex]	[latex]−2[/latex]
[latex]2[/latex]	[latex]−5[/latex]

 

 

 

 

 

 

 

 

 

Finally connect the points as best you can using a smooth curve.