10.5 Graphing quadratic equations (Page 4/15)

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Find the intercepts of the parabola $y = 9 x^{2} + 12 x + 4 .$

$y : (0, 4); x : (- \frac{2}{3}, 0)$

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: $(−1, −9)$ ;

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Graph the parabola $y = x^{2} - 8 x + 12 .$

$y : (0, 12); x : (2, 0), (6, 0);$
axis: $x = 4; vertex: (4, −4)$ ;

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Graph a quadratic equation in two variables.

Write the quadratic equation with $y$ on one side.
Determine whether the parabola opens upward or downward.
Find the axis of symmetry.
Find the vertex.
Find the y -intercept. Find the point symmetric to the y -intercept across the axis of symmetry.
Find the x -intercepts.
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 $x = - \frac{b}{2 a}$ .		The axis of symmetry is $x = 3 .$ The vertex is on the line $x = 3 .$
Find y when $x = 3 .$	The vertex is $(3, 0) .$
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 $y = 25 x^{2} + 10 x + 1 .$

$y : (0, 1); x : (- \frac{1}{5}, 0);$
axis: $x = - \frac{1}{5}; vertex: (- \frac{1}{5}, 0)$ ;
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 $x = - \frac{b}{2 a} .$	The axis of symmetry is $x = −2 .$
The vertex is on the line $x = −2 .$
Find y when $x = −2 .$	The vertex is $(−2, 1) .$
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 $(−4, 5) .$	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.
		$b^{2} - 4 a c$ $4^{2} - 4 \cdot 15$ $16 - 20$ $−4$
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: $x = \frac{3}{2}; vertex: (\frac{3}{2}, \frac{1}{2})$ ;
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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Binomial nomenclature

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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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