Learning Outcomes
- Graph quadratic equations of the form [latex]y=ax^2+bx+c[/latex]
- Identify important features of the graphs of quadratic equations
- Determine the maximum or minimum value of a quadratic equation
Keywords
- Parent equation: the simplest form of a general equation
- Parabola: the shape of any quadratic equation
- Vertex: the turning point of a parabola
- Line of symmetry: a line that cuts the graph into two mirror images
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.
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].
Examples
Determine the [latex]y[/latex]-intercept of the graph with equation:
1. [latex]y=4x^2-3x+6[/latex]
2. [latex]y=-2x^2-7[/latex]
3. [latex]y=x^2+5x[/latex]
Solution
1. Since [latex]c=6[/latex], the [latex]y[/latex]-intercept is the point [latex](0, 6)[/latex].
2. Since [latex]c=-7[/latex], the [latex]y[/latex]-intercept is the point [latex](0, -7)[/latex].
3. Since [latex]c=0[/latex], the [latex]y[/latex]-intercept is the point [latex](0, 0)[/latex].
Try It
Determine the [latex]y[/latex]-intercept of the graph with equation:
1. [latex]y=x^2-4x+1[/latex]
2. [latex]y=-2x^2-3[/latex]
3. [latex]y=x^2-6x[/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.
Example
Find the [latex]x[/latex]-intercepts for the graph of [latex]y={x}^{2}-x-6[/latex].
Solution
The [latex]x[/latex]-intercepts are found by solving the equation [latex]x^2 - x - 6 = 0[/latex].
| Solve. | [latex]x^2 - x - 6 = 0[/latex] |
| Factor. | [latex]\left( x - 3 \right) \left( x+2 \right) = 0[/latex] |
| Use the zero-factor property. | [latex]x - 3 = 0[/latex] and [latex]x +2 = 0[/latex] |
| Solve the resulting equations. | [latex]x=3[/latex] and [latex]x = -2[/latex] |
| The solutions are the [latex]x[/latex]-coordinate of the point. The [latex]y[/latex]-coordinate is 0. |
Answer
The [latex]x[/latex]-intercepts are [latex]\left(3, 0 \right)[/latex] and [latex] \left( -2, 0 \right) [/latex].
Try It
Find the [latex]x[/latex]-intercepts for the graph of [latex]y={x}^{2}-3x+2[/latex].
Try It
Find the [latex]x[/latex]-intercepts for the graph of [latex]y={x}^{2}-6x+9[/latex].
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.
Example
Find the [latex]x[/latex]-intercepts of the graph of [latex]y = x^2 + 4[/latex].
Solution
To find the [latex]x[/latex]-intercepts, we need to solve the equation [latex]x^2 + 4 = 0[/latex].
However, [latex]x^2 + 4[/latex] does not factor.
We can solve this equation by rewriting it:
[latex]x^2 + 4=0[/latex] can be rewritten as [latex]x^2 = -4 [/latex].
There are no real number values for [latex]x[/latex] that when squared result in [latex]-4[/latex].
The only value that [latex]x[/latex] can be are the imaginary numbers [latex]2i[/latex] and [latex]-2i[/latex] because [latex](2i)^2=4i^2=4(-1)=-4[/latex] and [latex](-2i)^2=4i^2=4(-1)=-4[/latex].
Since [latex]2i[/latex] and [latex]-2i[/latex] are complex numbers, they will not show up on the graph.
This parabola will have 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.
Try It
Find the [latex]x[/latex]-intercepts of the graph of [latex]y = 9x^2 + 1[/latex].
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.
Axis of SYmmetry and vertex
For the graph of the equation [latex]y=ax^2+bx+c[/latex], the axis of symmetry is the vertical line [latex]x=-\frac{b}{2a}[/latex].
The vertex is the point where [latex]x=-\frac{b}{2a}[/latex], paired with the corresponding [latex]y[/latex]-value.
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]
Example
Find the axis of symmetry and the vertex of the graph with equation [latex]y=-3x^2+x-2[/latex].
Solution
Axis: [latex]x=-\frac{b}{2a}=-\frac{1}{2\cdot (-3)}=-\frac{1}{-6}=\frac{1}{6}[/latex]
Vertex: [latex]x=\frac{1}{6}[/latex], so
[latex]\begin{equation}\begin{aligned}y&=-3x^2+x-2 \\ y&=-3{\left (\frac{1}{6}\right )}^{2}+\frac{1}{6}-2 \\ y&=-\frac{3}{1}\frac{1}{36}+\frac{1}{6}-2 \\ y&=-\frac{1}{12}+\frac{1}{6}-2 \\ y&=\frac{-1+2-24}{12} \\ &y=-\frac{23}{12}\end{aligned}\end{equation}[/latex]
Answer
The axis of symmetry is the vertical line [latex]x=\frac{1}{6}[/latex].
The vertex is the point [latex]\left (\frac{1}{6},-\frac{23}{12}\right )[/latex].
Try It
Find the axis of symmetry and the vertex of the graph of the equation [latex]y=2x^2+2x-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:
Graphs of Quadratic Functions
For [latex] \displaystyle y=a{{x}^{2}}+bx+c[/latex], where [latex]a, b[/latex] and [latex]c[/latex] are real numbers, and [latex]a\neq0[/latex],
- The parabola opens upward if [latex]a > 0[/latex] and downward if [latex]a < 0[/latex].
- The [latex]y[/latex]-intercept of the parabola occurs at the point [latex](0,c)[/latex].
- The [latex]x[/latex]-intercepts are found by solving the equation [latex]ax^2+bx+c=0[/latex]. The [latex]y[/latex]-coordinate is zero. There may be 0, 1, or 2 [latex]x[/latex]-intercepts.
- The axis of symmetry is the vertical line [latex]x=\frac{-b}{2a}[/latex].
- The vertex has an [latex]x[/latex]-coordinate of [latex]x=\dfrac{-b}{2a}[/latex]. The [latex]y[/latex]-coordinate is found by substituting this [latex]x[/latex]-value into the equation and solving for [latex]y[/latex].
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.
Example
Determine the maximum value of [latex]y[/latex]: [latex]y=-4x^2-4x+3[/latex]
Solution
Since the graph is a parabola that opens downwards, the maximum value will occur at the vertex.
Find the vertex: [latex]x=\frac{-b}{2a}=\frac{4}{2\cdot (-4)}=\frac{-1}{2}[/latex]
Then find the [latex]y[/latex]-value:
[latex]y=-4x^2-4x+3\\y=-4{\left (\frac{-1}{2}\right)}^{2}-4\left (\frac{-1}{2}\right )+3\\y=\frac{-4}{1}\cdot\frac{1}{4}+2+3\\y=-1+2+3\\y=4[/latex]
Answer
The maximum value is [latex]y=4[/latex].
Try It
Determine the minimum value of [latex]y[/latex]: [latex]y=x^2-8x-4[/latex]
