Graphing Systems of Inequalities, Part Two.

Graph the solution set of the following system

of inequalities. Here I have a system of nonlinear

inequalities. The first inequality is a quadratic

and the second is a line or a linear inequality.

We're first going to graph the top equation,

which is the quadratic inequality. So, I have

y is greater than or equal to x squared minus

9. In order to graph this inequality, we'll

rewrite the inequality sign with an equal

sign. Next, we're going to find the vertex.

To do that, we'll find negative b over 2a.

This is going to be the x-value of the vertex.

Now there is not a b term, because when we

have a quadratic the b term is the value in

front of the x. Since we don't have an x,

we don't have a b. So, that means this is

just 0 and it's being divided by 2 times 1,

the number in front of the x squared, and

0 divided by 2 is just 0. Now I'll find f

of 0. That means I'm just going to plug in

a 0 for x and solve for y. So, I have y equals

the 0 squared minus 9. So, y equals negative

9. So, my vertex is (0,-9).

If we plot this on our graph, we'll have a

point here at (0,-9) and the y-axis counts

by threes in this case. Now, I want to find

a value on the right and left of the vertex.

So, I'm going to choose x equals 1 and x equals

negative 1 and find ordered pairs. So, I just

created an XY chart. So, for x equals 1, I

have y equals 1 squared minus 9, which is

y equals 1 minus 9, which gives me a negative

8. So, this is (1,-8). Now, I have a negative

1 for x. So, I have negative 1 squared minus

9, which is 1 minus 9, which is also a negative

8. So, my second ordered pair is (-1,-8).

I went ahead and plotted those two points

on the graph and if we look at our inequality

sign we have the line underneath the inequality,

so I can go ahead and plot this quadratic

or draw this quadratic using a solid line.

Now I want to determine where to shade on

the graph. So, I'll choose a test point and

I'm going to choose (0,0) as my test point,

which is here. So, we'll plug in a 0 for y

and a 0 for x into the original inequality.

So, I have 0 is greater than or equal to 0

minus 9 and that 0 is greater than or equal

to a negative 9. That is true, so that means

point (0,0) is going to be included in my

graph.

So, I'll shade inside the parabola. Now that

I graphed my first inequality, I'm going to

graph the second inequality on the same coordinate

plane. My second inequality is a linear inequality.

So, I'm going to change the inequality sign

to an equal sign and then find the x-intercept.

To find the x-intercept, we'll just plug the

0 in for y. So, I get x minus 0 equals negative

3 and that tells me that x equals a negative

3. So, my x-intercept is (-3,0). Next, we'll

find our y-intercept. To do that, we're just

going to substitute a 0 for x and solve for

y.

So, I have 0 minus y equals a negative 3.

That gives me negative y equals negative 3

and I'll divide both sides by negative 1,

so I have y equals a positive 3. So, the y-intercept

is (0,3). We'll go to my graph and plot this

on the same plane. First, I want to graph

(-3,0), which is here, and then I'll plot

(0,3), which is here, and if we look at our

inequality sign, I have a line underneath

my sign, so I'm going to graph this with a

solid line. Now I want to decide to shade

above or below my line. So, I'll choose a

test point that is not on the line and I'm

going to choose point (0,0).

That means I'll substitute a 0 for x and 0

for y into the original inequality and I get

0 minus 0 is greater than or equal to negative

3, which is 0 is greater than or equal to

negative 3. That is a true statement, so that

means (0,0) is going to be included in my

graph. So, I'll shade beneath the line to

include that point. Now that I did both inequalities

on the same line and shaded them accordingly,

I'm going to only keep the shading of the

overlap of the two inequalities.

If we look at this graph, the overlap and

shading occurs in this region. So, the solution

set to the system of inequalities is shown

on the screen. I have the quadratic represented

by the black parabola and then the linear

inequality is represented by the blue line

and the overlap of their shading is represented

by the purple shading. So, this is my complete

answer to the problem.