We can write the boundary curve for Inequality (I) by replacing the less than sign with an equals sign. Then, we can identify which line it is.
x+y=4 ⇔ y=- x+4
Let's recall the equation of a line written in slope-intercept form.
y= mx+ b
Here, b is the y-intercept and m is the slope. We will rewrite the given equation to match this form, and then we can identify the characteristics of the slope.
y=- x+4 ⇔ y= - 1x+ 4
The y-intercept of the linear function is (0,4) and its slope is - 1. Let's draw its graph.
Now that we have the boundary line, 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 strict inequality, the boundary line will be dashed.
Inequality (II)
We can write the boundary curve by replacing the greater than or equal to sign with an equals sign.
9x^2-4y^2=36
To identify which conic section it is we need to rewrite the equation.
Let's recall the equation of hyperbolas centered at the origin.
ccc
Horizontal & & Vertical
Hyperbola & & Hyperbola
x^2/a^2-y^2/b^2=1 & & y^2/a^2-x^2/b^2=1
From the above formula, we can see that our equation represents a horizontal hyperbola. Let's review the main characteristics of this type of hyperbola.
Using this information, we can identify that the vertices are (± 2 ,0). Let's substitute a=2 and b=3 into the formula for the asymptotes and obtain their equations.
y=± b/ax ⇒ y=± 3/2x
The asymptotes are y=± 32x. Now, let's calculate c, the absolute value of the nonzero coordinate of the foci. To do so, we will substitute a=2 and b=3 into c^2=a^2+b^2.
The foci of the hyperbola are (± sqrt(13),0). To graph the function, let's summarize all of the information that we have found.
Equation
x^2/2^2-y^2/3^2=1
Transverse axis
Horizontal
Vertices
(± 2,0)
Foci
(± sqrt(13),0)
Asymptotes
y=± 3/2x
Let's now draw the hyperbola. Because the inequality is non-strict, the boundary curve will be solid.
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.
