y≥ x^2-4 & (I) y<-3 x-1 & (II)
Let's graph each of them, one at a time.
Inequality (I)
To graph the inequality, we have to draw the boundary curve. We can do it by replacing the greater than or equal to sign with an equals sign. Since it is a quadratic inequality, let's identify the coefficients a, b, and c.
y=x^2-4 ⇕ y= 1x^2+ 0x+( -4)
Knowing that a= 1, b= 2, and c= 0, we can find the vertex of the parabola. To do so we will need to think of y as a function of x, y=f(x).
Vertex of the Parabola ( - b/2a,f(- b/2a) )Let's substitute the values of a and b in the formula for the x-coordinate of the vertex.
The vertex is (0, -4). With this we know that the axis of symmetry of the parabola is the vertical line x=-4. Next, let's find two more points on the curve — one on each side of the axis of symmetry.
x
x^2-4
y=x^2-4
-2
( -2)^2-4
0
2
2^2-4
0
The points ( -2, 0) and ( 2, 0) are on the parabola. Let's plot the points and connect them with a smooth curve.
Now that we have the boundary curve, we need to determine which region to shade. To do so we will use (0,0) as a test point. If the point satisfies the inequality, we will shade the region that contains the point. If not, we will shade the opposite region.
Since the substitution produced a true statement, we will shade the region that contains the point (0,0). Because we have a non-strict inequality, the boundary curve will be solid.
Inequality (II)
Once again, we can write the boundary curve by replacing the greater than sign with an equals sign.
ccc
Inequality & &Boundary Line
y > -3 x+1 & & y = -3 x+1
Fortunately, this equation is already in slope-intercept form, so we can identify the slope m and y-intercept (0, b).
y= -3x+ 1
We will plot the y-intercept (0, 1), then use the slope m= -3 to plot another point on the line. Connecting these points with a dashed line will give us the boundary line of our inequality.
Note that the boundary line is dashed, because the inequality is strict. To decide which side of the boundary line to shade, we will substitute a test point that is not on the boundary line into the given inequality. Let's use (0,0) as our test point.
Since the substitution of the test point created a true statement, we will shade the region that contains the point.
Solution
The solution set is the overlapping region.
b Similarly, we can graph the second system.
y<2x+5 & (I) y≥ |x+1| & (II)
Let's begin with Inequality (I).
Inequality (I)
We can see that the inequality is linear. To obtain the equation of the boundary line, we should replace the inequality with an equals sign.
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Inequality &
&Boundary Line
y < 2x+5 &
& y = 2x+5
Since this equation is already in slope-intercept form, so we can identify the slope m and y-intercept (0, b).
y= 2x+ 5
We will plot the y-intercept (0, 5), then use the slope m= 2 to plot another point on the line. The boundary line is dashed because the inequality is strict.
Now, to decide which side of the boundary line to shade we will substitute a test point that is not on the boundary line into the given inequality. Let's use (0,0) as our test point.
We received a true statement, so we will shade the region that contains the point.
Inequality (II)
Once again, let's begin by exchanging the inequality symbol for an equals sign.
ccc
Inequality &
&Boundary Line
y ≥ |x+1| &
& y = |x+1|
The graph of this boundary line is the graph of the parent function y=|x| translated 1 unit left. The boundary line will be solid because the inequality is non-strict.
Next, we need to decide which side of the boundary line we should shade. Let's use (0,0).
