McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
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Exercise 73 Page 650

Both inequalities are strict. What does it mean in terms of the graph?

Practice makes perfect

We want to solve the given system of inequalities by graphing. Note that both inequalities of the system are quadratic inequalities. x^2+y^2<36 & (I) 4x^2+9y^2>36 & (II) Let's graph each of them, one at a time.

Inequality (I)

We can write the boundary curve for Inequality (I) by replacing the less than sign with an equals sign. Then, we can identify which conic section it is. x^2+y^2=36 Let's recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 Here, the center is the point ( h, k) and the radius is r. We will rewrite the given equation to match this form, and then we can identify the center and the radius. x^2+y^2=36 [0.3em] ⇕ [0.3em] (x- 0)^2+(y- 0)^2= 6^2 The center of the circle is the point ( 0, 0), and its radius is 6. Let's draw its graph.

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.

x^2+y^2<36
0^2+ 0^2? <36
â–¼
Simplify left-hand side
0+0? <36
0<36

Since the substitution produced a true statement, we will shade the region that contains the point (0,0). Because we have a strict inequality, the boundary curve will be dashed.

Inequality (II)

We can write the boundary curve by replacing the greater than sign with an equals sign. 4x^2+9y^2=36 To identify which conic section it is we need to rewrite the equation.

4x^2+9y^2=36
â–¼
Simplify
4x^2/36+9y^2/36=1
x^2/9+y^2/4=1
x^2/3^2+y^2/2^2=1
(x-0)^2/3^2+(y-0)^2/2^2=1

Notice that the denominator of the expression containing the x-variable is greater than the denominator of the expression that contains the y-variable. Therefore, we have a horizontal ellipse. Let's recall the main characteristics of this type of ellipse.

Horizontal Ellipse
Standard-Form Equation (x- h)^2/a^2+(y-k)^2/b^2=1
Center ( h,k)
Vertices ( h± a,k)
Co-vertices ( h,k ± b)
Foci ( h± c,k)
a,b,c relationship, a>b>0 c^2= a^2- b^2

Let's consider our equation one more time. (x- 0)^2/3^2+(y- )^2/2^2=1 We can see that a= 3, b= 2, h= 0, and k= . The only value we do not know is c. Let's find it!

c^2=a^2-b^2
c^2= 3^2- 2^2
â–¼
Solve for c
c^2=9-4
c^2=5
c=± sqrt(5)

Let's now summarize the properties of the ellipse.

Horizontal Ellipse
Standard-Form Equation (x- 0)^2/3^2+(y- )^2/2^2=1
Center ( 0, )
Vertices ( 0± 3, )
⇓
(3,0 )and(- 3,0 )
Co-vertices ( 0, ± 2)
⇓
(0,2 )and(0,- 2 )
Foci ( 0± sqrt(5), )
⇓
(sqrt(5),0 )and(- sqrt(5),0 )
c sqrt(5)

Let's now draw the ellipse on the same set of axis as we drew the circle. Because the inequality is strict, the boundary curve will be dashed.

Now that we have the boundary curve, we need to determine which region to shade. Again, we will use (0,0) as a test point. Let's determine if the point satisfies the inequality or not.

4x^2+9y^2>36
4( 0)^2+9( 0)^2? >36
â–¼
Simplify left-hand side
0+0? >36
0≯ 36

Since the substitution produced a false statement, we will shade the region that does not contain the point (0,0).

Solution

The solution set is the overlapping region.