MAT 1302 Unit VII-Graphing Systems of Inequalities Part II Q14

10 days ago

English subtitle

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.