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Find the intercepts of the parabola $y=9{x}^{2}+12x+4.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,4\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{2}{3},0\right)$
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.
Graph the parabola $y={x}^{2}+2x-8.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,\mathrm{-8}\right)$ ;
$x\text{:}\phantom{\rule{0.2em}{0ex}}\left(2,0\right),\left(\mathrm{-4},0\right)$ ;
axis:
$x=\mathrm{-1}$ ; vertex:
$\left(\mathrm{-1},\mathrm{-9}\right)$ ;
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:
$x=4;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(4,\mathrm{-4}\right)$ ;
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 $y=\text{\u2212}{x}^{2}+6x-9$ .
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 $(3,0).$ | |
The
y -intercept occurs when
$x=0.$
Substitute $x=0.$ Simplify. The point $(0,\mathrm{-9})$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $(6,\mathrm{-9}).$ Point symmetric to the y- intercept is $(6,\mathrm{-9})$ |
The y -intercept is $(0,\mathrm{-9}).$ | |
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. |
Graph the parabola $y=\mathrm{-3}{x}^{2}+12x-12.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,\mathrm{-12}\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(2,0\right);$
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 $0=\text{\u2212}{x}^{2}+6x-9$ 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 $y={x}^{2}+4x+5$ .
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 $x=\mathrm{-2}.$ | |
The vertex is on the line $x=\mathrm{-2}.$ | ||
Find y when $x=\mathrm{-2}.$ |
The vertex is $(\mathrm{-2},1).$ | |
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 $(\mathrm{-4},5).$ |
The y -intercept is $(0,5).$ Point symmetric to the y- intercept is $(\mathrm{-4},5)$ . | |
The x - intercept occurs when $y=0.$ | ||
Substitute
$y=0.$
Test the discriminant. |
| |
${b}^{2}-4ac$
${4}^{2}-4\cdot 15$ $16-20$ $\phantom{\rule{1em}{0ex}}\mathrm{-4}$ | ||
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 $y=\mathrm{-2}{x}^{2}-1.$
$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,\mathrm{-1}\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis:
$x=0;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(0,\mathrm{-1}\right)$ ;
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