How do you graph #(x^2-4)/(x^2-9)#?

1. Disk Graph 2 1 3 X 48
2. Disk Graph 2 1 3 X 4
3. Disk Graph 2 1 3 X 4 5
4. Disk Graph 2 1 3 X 4

1 Answer

The shape is revolved about the line x = 4. The boundaries of the shape being revolved are y = x 3, y = 0, and x = 2. If you draw a diagram, you should see that beween using disks or cylindrical shells, the shells are the best choice. The height of the region being revolved at position x is x 3. And that position is a distance 4 - x from the axis. Graph 2: These are your 'Least' responses for each of the four scales. Graph 3: This graph is the result of combining your 'Most' choices with your 'Least' choices and is used to determine your highest DiSC dimension, Intensity Index scores, and Classical Profile Pattern. (Graph 3 is automatically included in the Online DiSC Classic 2.0 and DiSC Classic 2 Plus Reports).

You find the intercepts and the asymptotes, and then you sketch the graph.

Explanation:

Step 1. Find the #y#-intercepts.

#y = f(x) = (x^2-4)/(x^2-9)#

#f(0) = (0^2-4)/(0^2-9) = (-4)/-9 = 4/9#

Disk Graph 2 1 3 X 48

The #y#-intercept is at (#0,4/9#).

Step 2. Find the x-intercepts.

#0 = (x^2-4)/(x^2-9)#

#x^2-4 = 0#

#x^2 = 4#

#x = ±2#

The #x#-intercepts are at (#-2,0#) and (#2,0#).

Step 3. Find the vertical asymptotes.

Set the denominator equal to zero and solve for #x#.

#x^2 -9 = 0#

#x^2 = 9#

#x = ±3#

There are vertical asymptotes at #x = -3# and #x = 3#.

Step 4. Find the horizontal asymptote.

Both equations are of the second order, so we divide the coefficients of the #x^2# terms.

#1/1 = 1#

The horizontal asymptote is at #y=1#.

Step 5. Draw your axes and the asymptotes.

The vertical asymptotes divide the graph into three regions of #x#s.

Step 6. Sketch the graph in the each region.

(a) In the left hand region,

#f(-4) = (16-4)/(16-9) = 12/7 ≈ 1.7#.

The point at (#-4,1.7#) is in the second quadrant, so we have a 'hyperbola' above the horizontal asymptote.

(b) In the right hand region,

#f(4) = (16-4)/(16-9) = 12/7 ≈ 1.7#.

So we have a mirror-image 'hyperbola' in the first quadrant.

(c) In the middle region, we have

#f(0) = 4/9 ≈ 0.44# and

Disk Graph 2 1 3 X 4

#f(-1) = f(1) = (1-4)/(1-9) =(-3)/-8 ≈ 0.38#

The points at (#-1,0.38#) and (#1,0.38#) are below the #y#-intercept, so we have an 'inverted parabola' between the vertical asymptotes.

And we have our graph.

Disk Graph 2 1 3 X 4 5

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