# 10.5 Graphing quadratic equations  (Page 4/15)

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,4\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{2}{3},0\right)$

## 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 $y={x}^{2}-6x+8$ .

## Solution

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-8\right)$ ; $x\text{:}\phantom{\rule{0.2em}{0ex}}\left(2,0\right),\left(-4,0\right)$ ;
axis: $x=-1$ ; vertex: $\left(-1,-9\right)$ ;

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,12\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(2,0\right),\left(6,0\right);$
axis: $x=4;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(4,-4\right)$ ;

## 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 $y=\text{−}{x}^{2}+6x-9$ .

## Solution

 The equation y has on one side. Since a is $-1$ , the parabola opens downward. To find the axis of symmetry, find $x=-\frac{b}{2a}$ . The axis of symmetry is $x=3.$ The vertex is on the line $x=3.$ Find y when $x=3.$ The vertex is $\left(3,0\right).$ The y -intercept occurs when $x=0.$ Substitute $x=0.$ Simplify. The point $\left(0,-9\right)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $\left(6,-9\right).$ Point symmetric to the y- intercept is $\left(6,-9\right)$ The y -intercept is $\left(0,-9\right).$ 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.

Graph the parabola $y=-3{x}^{2}+12x-12.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-12\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(2,0\right);$
axis: $x=2;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\left(2,0\right)$ ;

Graph the parabola $y=25{x}^{2}+10x+1.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,1\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{1}{5},0\right);$
axis: $x=-\frac{1}{5};\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{1}{5},0\right)$ ;

For the graph of $y=-{x}^{2}+6x-9$ , 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 $0=\text{−}{x}^{2}+6x-9$ 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 $y={x}^{2}+4x+5$ ?

Graph $y={x}^{2}+4x+5$ .

## Solution

 The equation has y on one side. Since a is 1, the parabola opens upward. To find the axis of symmetry, find $x=-\frac{b}{2a}.$ The axis of symmetry is $x=-2.$ The vertex is on the line $x=-2.$ Find y when $x=-2.$ The vertex is $\left(-2,1\right).$ The y -intercept occurs when $x=0.$ Substitute $x=0.$ Simplify. The point $\left(0,5\right)$ is two units to the right of the line of symmetry. The point two units to the left of the line of symmetry is $\left(-4,5\right).$ The y -intercept is $\left(0,5\right).$ Point symmetric to the y- intercept is $\left(-4,5\right)$ . The x - intercept occurs when $y=0.$ Substitute $y=0.$ Test the discriminant. ${b}^{2}-4ac$ ${4}^{2}-4\cdot 15$ $16-20$ $\phantom{\rule{1em}{0ex}}-4$ 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.

Graph the parabola $y=2{x}^{2}-6x+5.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,5\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=\frac{3}{2};\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(\frac{3}{2},\frac{1}{2}\right)$ ;

Graph the parabola $y=-2{x}^{2}-1.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-1\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=0;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(0,-1\right)$ ;