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10.5 Graphing quadratic equations  (Page 4/15)

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Find the intercepts of the parabola y = 9 x 2 + 12 x + 4 .

y : ( 0 , 4 ) ; x : ( − 2 3 , 0 )

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Graph quadratic equations in two variables

Now, we have all the pieces we need in order to graph a quadratic equation in two variables. We just need to put them together. In the next example, we will see how to do this.

How to graph a quadratic equation in two variables

Graph y = x 2 − 6 x + 8 .

Solution

The image shows the steps to graph the quadratic equation y equals x squared minus 6 x plus 8. Step 1 is to write the quadratic equation with y on one side. This equation has y on one side already. The value of a is one, the value of b is -6 and the value of c is 8. Step 2 is to determine whether the parabola opens upward or downward. Since a is positive, the parabola opens upward. Step 3 is to find the axis of symmetry. The axis of symmetry is the line x equals negative b divided by the quantity 2 a. Plugging in the values of b and a the formula becomes x equals negative -6 divided by the quantity 2 times 1 which simplifies to x equals 3. The axis of symmetry is the line x equals 3. Step 4 is to find the vertex. The vertex is on the axis of symmetry. Substitute x equals 3 into the equation and solve for y. The equation is y equals x squared minus 6 x plus 8. Replacing x with 3 it becomes y equals 3 squared minus 6 times 3 plus 8 which simplifies to y equals -1. The vertex is (3, -1). Step 5 is to find the y-intercept and find the point symmetric to the y-intercept across the axis of symmetry. We substitute x equals 0 into the equation. The equation is y equals x squared minus 6 x plus 8. Replacing x with 0 it becomes y equals 0 squared minus 6 times 0 plus 8 which simplifies to y equals 8. The y-intercept is (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x equals 6. The point symmetric to the y-intercept is (6, 8). Step 6 is to find the x-intercepts. We substitute y equals 0 into the equation. The equation becomes 0 equals x squared minus 6 x plus 8. We can solve this quadratic equation by factoring to get 0 equals the quantity x minus 2 times the quantity x minus 4. Solve each equation to get x equals 2 and x equals 4. The x-intercepts are (2, 0) and (4, 0). Step 7 is to graph the parabola. We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these five points to sketch the parabola. The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -2 to 10. The y-axis of the plane runs from -3 to 10. The vertex is at the point (3, -1). Four points are plotted on the curve at (0, 8), (6, 8), (2, 0) and (4, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3.
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Graph the parabola y = x 2 + 2 x − 8 .

y : ( 0 , −8 ) ; x : ( 2 , 0 ) , ( −4 , 0 ) ;
axis: x = −1 ; vertex: ( −1 , −9 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (-1, -9). Three points are plotted on the curve at (0, -8), (2, 0) and (-4, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1.

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Graph the parabola y = x 2 − 8 x + 12 .

y : ( 0 , 12 ) ; x : ( 2 , 0 ) , ( 6 , 0 ) ;
axis: x = 4 ; vertex: ( 4 , −4 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (4, -4). Three points are plotted on the curve at (0, 12), (2, 0) and (6, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 4.

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Graph a quadratic equation in two variables.

  1. Write the quadratic equation with y on one side.
  2. Determine whether the parabola opens upward or downward.
  3. Find the axis of symmetry.
  4. Find the vertex.
  5. Find the y -intercept. Find the point symmetric to the y -intercept across the axis of symmetry.
  6. Find the x -intercepts.
  7. Graph the parabola.

We were able to find the x -intercepts in the last example by factoring. We find the x -intercepts in the next example by factoring, too.

Graph y = − x 2 + 6 x − 9 .

Solution

The equation y has on one side. .
Since a is − 1 , the parabola opens downward.

To find the axis of symmetry, find x = − b 2 a .		 . .
.
.
The axis of symmetry is x = 3 . The vertex is on the line x = 3 .
.
Find y when x = 3 . .
.
.
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The vertex is ( 3 , 0 ) .
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The y -intercept occurs when x = 0 .
Substitute x = 0 .
Simplify.

The point ( 0 , −9 ) is three units to the left of the line of symmetry.
The point three units to the right of the line of symmetry is ( 6 , −9 ) .
Point symmetric to the y- intercept is ( 6 , −9 ) 		.
.
.
The y -intercept is ( 0 , −9 ) .
.
The x -intercept occurs when y = 0 . .
Substitute y = 0 . .
Factor the GCF. .
Factor the trinomial. .
Solve for x . .
Connect the points to graph the parabola. .
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Graph the parabola y = −3 x 2 + 12 x − 12 .

y : ( 0 , −12 ) ; x : ( 2 , 0 ) ;
axis: x = 2 ; vertex: ( 2 , 0 ) ;
The graph shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -1 to 10. The vertex is at the point (2, 0). One other point is plotted on the curve at (0, -12). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2.

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Graph the parabola y = 25 x 2 + 10 x + 1 .

y : ( 0 , 1 ) ; x : ( − 1 5 , 0 ) ;
axis: x = − 1 5 ; vertex: ( − 1 5 , 0 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 10. The vertex is at the point (-1 fifth, 0). One other point is plotted on the curve at (0, 1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1 fifth.

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For the graph of y = − x 2 + 6 x − 9 , the vertex and the x -intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation 0 = − x 2 + 6 x − 9 is 0, so there is only one solution. That means there is only one x -intercept, and it is the vertex of the parabola.

How many x -intercepts would you expect to see on the graph of y = x 2 + 4 x + 5 ?

Graph y = x 2 + 4 x + 5 .

Solution

The equation has y on one side. .
Since a is 1, the parabola opens upward. .
To find the axis of symmetry, find x = − b 2 a . .
.
.
The axis of symmetry is x = −2 .
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The vertex is on the line x = −2 .
Find y when x = −2 . .
.
.
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The vertex is ( −2 , 1 ) .
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The y -intercept occurs when x = 0 .
Substitute x = 0 .
Simplify.
The point ( 0 , 5 ) is two units to the right of the line of symmetry.
The point two units to the left of the line of symmetry is ( −4 , 5 ) . 		.
.
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The y -intercept is ( 0 , 5 ) .
.
Point symmetric to the y- intercept is ( −4 , 5 ) .
The x - intercept occurs when y = 0 .
Substitute y = 0 .
Test the discriminant.		 .
.
b 2 − 4 a c
4 2 − 4 ⋅ 1 5
16 − 20
−4
Since the value of the discriminant is negative, there is no solution and so no x- intercept.
Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.		 .
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Graph the parabola y = 2 x 2 − 6 x + 5 .

y : ( 0 , 5 ) ; x : none ;
axis: x = 3 2 ; vertex: ( 3 2 , 1 2 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 10. The vertex is at the point (3 halves, 1 half). One other point is plotted on the curve at (0, 5). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3 halves.

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Graph the parabola y = −2 x 2 − 1 .

y : ( 0 , −1 ) ; x : none ;
axis: x = 0 ; vertex: ( 0 , −1 ) ;
The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (0, -1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 0.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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