10.5: Graphing Quadratic Equations

Learning Objectives
Be Prepared

Before you get started, take this readiness quiz.

  1. Graph the equation \(y=3x−5\) by plotting points.
    If you missed this problem, review [link].
  2. Evaluate \(2x^2+4x−1\)when \(x=−3\)
    If you missed this problem, review [link].
  3. Evaluate \(−\frac\) when \(a=13\) and b=\(\frac\)
    If you missed this problem, review [link].

Recognize the Graph of a Quadratic Equation in Two Variables

We have graphed equations of the form \(Ax+By=C\). We called equations like this linear equations because their graphs are straight lines.

Now, we will graph equations of the form \(y=ax^2+bx+c\). We call this kind of equation a quadratic equation in two variables.

definition: QUADRATIC EQUATION IN TWO VARIABLES

A quadratic equation in two variables, where a,b,and c are real numbers and \(a\neq 0\), is an equation of the form \[y=ax^2+bx+c \nonumber\]

Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations.

Let’s look first at graphing the quadratic equation \(y=x^2\). We will choose integer values of x between −2 and 2 and find their y values. See Table.

one in the x column and the number one in the y column. The fifth row has the number negative one in the x column and the number one in the y column. The sixth row has the number two in the x column and the number four in the y column. The last row has the number negative two in the x column and the number four in the y column.">
\(y=x^2\)
x y
0 0
1 1
\(−1\) 1
2 4
\(−2\) 4

Notice when we let \(x=1\) and \(x=−1\), we got the same value for y.

The same thing happened when we let \(x=2\) and \(x=−2\).

Now, we will plot the points to show the graph of \(y=x^2\). See Figure.

This figure shows an upward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (0, 0). Other points on the curve are located at (-2, 4), (-1, 1), (1, 1) and (2, 4).

The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this.

In Example you will practice graphing a parabola by plotting a few points.

Example \(\PageIndex\)

Answer

We will graph the equation by plotting points.

one in the y column. The fourth row has the number one in the x column and the number zero in the y column. The fifth row has the number negative one in the x column and the number zero in the y column. The sixth row has the number two in the x column and the number three in the y column. The last row has the number negative two in the x column and the number three in the y column. Step 3 is to plot the points, and then connect them with a smooth curve. The graph of the curve is an upward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (0, -1). Other points on the curve are located at ( -2, 3), (-1, 0), (1, 0) and (2, 3).">

Choose integers values for x, substitute them into the equation and solve for y.
Record the values of the ordered pairs in the chart. .
Plot the points, and then connect them with a smooth curve. The result will be the graph of the equation \(y=x^2−1\) .
Example \(\PageIndex\)

Answer

This figure shows a downward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (0, 0). Other points on the curve are located at (-2, -4), (-1, -1), (1, -1) and (2, -4).

Example \(\PageIndex\)

Answer

This figure shows an upward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (0, 1). Other points on the curve are located at (-2, 5), (-1, 2), (1, 2) and (2, 5).

How do the equations \(y=x^2\) and \(y=x^2−1\) differ? What is the difference between their graphs? How are their graphs the same?

All parabolas of the form \(y=ax^2+bx+c\) open upwards or downwards. See Figure .

This figure shows two graphs side by side. The graph on the left side shows an upward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Below the graph is the equation y equals a squared plus b x plus c. Below that is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the inequality a greater than 0 which means the parabola opens upwards. The graph on the right side shows a downward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Below the graph is the equation y equals a squared plus b x plus c. Below that is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the inequality a less than 0 which means the parabola opens downwards.

Notice that the only difference in the two equations is the negative sign before the \(x^2\) in the equation of the second graph in Figure. When the \(x^2\) term is positive, the parabola opens upward, and when the \(x^2\) term is negative, the parabola opens downward.

Definition: PARABOLA ORIENTATION

For the quadratic equation \(y=ax^2+bx+c\), if:

The image shows two statements. The first statement reads “a greater than 0, the parabola opens upwards”. This statement is followed by the image of an upward opening parabola. The second statement reads “a less than 0, the parabola opens downward”. This statement is followed by the image of a downward opening parabola.

Example\(\PageIndex\)

Determine whether each parabola opens upward or downward:

  1. \(y=−3x^2+2x−4\)
  2. \( y=6x^2+7x−9\)

.

Since the “a” is negative, the parabola will open downward.

.

Since the “a” is positive, the parabola will open upward.

Example\(\PageIndex\)

Determine whether each parabola opens upward or downward:

Example \(\PageIndex\)

Determine whether each parabola opens upward or downward:

Find the Axis of Symmetry and Vertex of a Parabola

Look again at Figure. Do you see that we could fold each parabola in half and that one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.

We show the same two graphs again with the axis of symmetry in red. See Figure.

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (-2, -1). Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (2, 7). Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3.

The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of \(y=ax^2+bx+c\) is x=\(−\frac\).

So, to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula x=\(−\frac\).

The figure shows the steps to find the axis of symmetry for two parabolas. On the left side the standard form of a quadratic equation which is y equals a x squared plus b x plus c is written above the given equation y equals x squared plus 4 x plus 3. The axis of symmetry is the equation x equals negative b divided by the quantity two times a. Plugging in the values of a and b from the quadratic equation the formula becomes x equals negative 4 divided by the <a href=quantity 2 times 1, which simplifies to x equals negative 2. On the right side the standard form of a quadratic equation which is y equals a x squared plus b x plus c is written above the given equation y equals negative x squared plus 4 x plus 3. The axis of symmetry is the equation x equals negative b divided by the quantity two times a. Plugging in the values of a and b from the quadratic equation the formula becomes x equals negative 4 divided by the quantity 2 times -1, which simplifies to x equals 2." />

The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards. This point is called the vertex of the parabola.

We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its x-coordinate is \(−\frac\). To find the y-coordinate of the vertex, we substitute the value of the x-coordinate into the quadratic equation.

The figure shows the steps to find the vertex for two parabolas. On the left side is the given equation y equals x squared plus 4 x plus 3. Below the equation is the statement “axis of symmetry is x equals -2”. Below that is the statement “vertex is” next to the statement is an ordered pair with x-value of -2, the same as the axis of symmetry, and the y-value is blank. Below that the original equation is rewritten. Below the equation is the equation with -2 plugged in for the x value which is y equals -2 squared plus 4 times -2 plus 3. This simplifies to y equals -1. Below this is the statement “vertex is (-2, -1)”. On the right side is the given equation y equals negative x squared plus 4 x plus 3. Below the equation is the statement “axis of symmetry is x equals 2”. Below that is the statement “vertex is” next to the statement is an ordered pair with x-value of 2, the same as the axis of symmetry, and the y-value is blank. Below that the original equation is rewritten. Below the equation is the equation with 2 plugged in for the x value which is y equals negative the quantity 2 squared, plus 4 times 2 plus 3. This simplifies to y equals 7. Below this is the statement “vertex is (2, 7)”.

Definition: AXIS OF SYMMETRY AND VERTEX OF A PARABOLA

For a parabola with equation \(y=ax^2+bx+c\):

To find the y-coordinate of the vertex, we substitute x=\(−\frac\) into the quadratic equation.

Example\(\PageIndex\)

For the parabola \(y=3x^2−6x+2\) find:

  1. the axis of symmetry and
  2. the vertex.
quantity 2 times 3, Step 2 is to simplify. The equation simplifies to x equals 1. The answer is “The axis of symmetry is the line x equals 1”. Part b shows the given equation, y equals 3 x squared minus 6 x plus 2 The vertex is on the line on symmetry, so its x coordinate will be x equals 1. Step 1 is to substitute x equal 1 into the equation and solve for y. The equation becomes y equals 3 times 1 squared minus 6 times 1 plus 2. Step 2 is to simplify the equation. The equation simplifies to y equals 3 times 1 minus 6 plus 2 which is -1. Step three is to write the ordered pair. The vertex is (1, -1).">
1. .
The axis of symmetry is the line x=\(−\frac\) .
Substitute the values of a, b into the equation. .
Simplify x=1
The axis of symmetry is the line x=1
2. .
The vertex is on the line of symmetry, so its x-coordinate will be x=1
Substitute x=1 into the equation and solve for y. .
Simplify .
This is the y-coordinate. y=−1
The vertex is (1,−1).
Example \(\PageIndex\)

For the parabola \(y=2x^2−8x+1\) find:

Example \(\PageIndex\)

For the parabola \(y=2x^2−4x−3\) find:

Find the Intercepts of a Parabola

When we graphed linear equations, we often used the x- and y-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y-intercept the value of x is zero. So, to find the y-intercept, we substitute x=0 into the equation.

Let’s find the y-intercepts of the two parabolas shown in the figure below.

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2. Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the statement “x equals 0”. Next to that is the equation of the graph with 0 plugged in for x which gives y equals 0 squared plus4 times 0 plus 3. This simplifies to y equals 3. Below the equation is the statement “y-intercept (0, 3)”. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2. Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the statement “x equals 0”. Next to that is the equation of the graph with 0 plugged in for x which gives y equals negative quantity 0 squared plus 4 times 0 plus 3. This simplifies to y equals 3. Below the equation is the statement “y-intercept (0, 3)”.

At an x-intercept, the value of y is zero. To find an x-intercept, we substitute \(y=0\) into the equation. In other words, we will need to solve the equation \(0=ax^2+bx+c\) for x.

But solving quadratic equations like this is exactly what we have done earlier in this chapter.

We can now find the x-intercepts of the two parabolas shown in Figure.

First, we will find the x-intercepts of a parabola with equation \(y=x^2+4x+3\).

.
Let y=0 .
Factor. .
Use the zero product property. .
Solve. .
The x intercepts are (−1,0) and (−3,0).

Now, we will find the x-intercepts of the parabola with equation \(y=−x^2+4x+3\).

.
Let y=0 .
This quadratic does not factor, so we use the Quadratic Formula. .
a=−1, b=4, c=3. .
Simplify. .
.
..
The x intercepts are \((2+\sqrt,0)\) and \((2−\sqrt,0)\)

We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph.

Do these results agree with our graphs? See Figure.

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (-2, -1). Three points are plotted on the curve at (-3, 0), (-1, 0), and (0, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2. Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the statement “y-intercept (0, 3)”. Below that is the statement “x-intercepts (-1, 0) and (-3, 0)”. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The vertex is at the point (2, 7). Three points are plotted on the curve at (-0.6, 0), (4.6, 0), and (0, 3). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2. Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the statement “y-intercept (0, 3)”. Below that is the statement “x-intercepts (2 plus square root of 7, 0) is approximately equal to (4.6, 0) and (2 minus square root of 7, 0) is approximately equal to (-0.6, 0).”

Definition: FIND THE INTERCEPTS OF A PARABOLA

To find the intercepts of a parabola with equation \(y=ax^2+bx+c\):

Example \(\PageIndex\)

Find the intercepts of the parabola \(y=x^2−2x−8\).

Answer

.

When y=0, then x=4 or x=−2. The x-intercepts are the points (4,0) and (−2,0).
Example \(\PageIndex\)

Find the intercepts of the parabola \(y=x^2+2x−8\).

Answer

Example \(\PageIndex\)

Find the intercepts of the parabola \(y=x^2−4x−12\).

Answer

In this chapter, we have been solving quadratic equations of the form \(ax^2+bx+c=0\). We solved for xx and the results were the solutions to the equation.

We are now looking at quadratic equations in two variables of the form \(y=ax^2+bx+c\). The graphs of these equations are parabolas. The x-intercepts of the parabolas occur where y=0.

The solutions of the quadratic equation are the x values of the x-intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the equations give the x-intercepts of the graphs, the number of x-intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic equation of the form \(ax^2+bx+c=0\). Now, we can use the discriminant to tell us how many x-intercepts there are on the graph.

This figure shows three graphs side by side. The leftmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. Below the graph is the inequality b squared minus 4 a c greater than 0. Below that is the statement “Two solutions”. Below that is the statement “ Two x-intercepts”. The middle graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is on the x-axis. Below the graph is the equation b squared minus 4 a c equals 0. Below that is the statement “One solution”. Below that is the statement “ One x-intercept”. The rightmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper left quadrant. Below the graph is the inequality b squared minus 4 a c less than 0. Below that is the statement “No real solutions”. Below that is the statement “ No x-intercept”.

Before you start solving the quadratic equation to find the values of the x-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Example \(\PageIndex\)

Find the intercepts of the parabola \(y=5x^2+x+4\).

Answer