Could you help me sketch the curve: f(x) = [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2) ]?

Question

Could you help me sketch the above curve.
Also, what is the domain, what is the range, is there any symmetry, x-intercepts, y-intercepts, horizontal and vertical asymptotes, undefined points, holes in the graph, and what would be the best process, the best order of items to check for, and the best approach to graphing this and other rational functions? Thanks for your help.

Answer 1

f(x) = [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2) ]
= 3(x - 2)/(x - 3) (x ≠2, -2)
= [3(x - 3) + 3]/(x - 3)
= 3 + 3/(x - 3)

1. Lim x→2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]
= lim x→2 [3 + 3/(x - 3)]
= 0
and
Lim x→-2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]
= lim x→-2 [3 + 3/(x - 3)]
= 13/5
So it has 'holes' at (2, 0) and (-2, 13/5)

2. It is undefined when x = 3, 2 and -2
So domain is all x except x = 3, 2, -2
Range all y except y = 0 and 13/5

3. As x increases and decreases without bound
3/(x-3) approaches 0
so 3 + 3/(x-3) approaches 3

ie lim x → ±∞ [3 + 3/(x-3)] = 3

4. When x < 3
x - 3 < 0
So 3 + 3/(x-3) < 3

When x > 3
x - 3 > 0
So 3 + 3/(x-3) > 3

So for x > 3 the curve is above y = 3 and for x< 3 the curve is below y = 3

5. When y = f(x) = 0
3(x - 2)/(x - 3) = 0
ie x = 2
So f(x) cuts x-axis at x = 2

6. When x = 0
f(x) = 3*(-2)/(-3) = 2
So y-intercept is y = 2

7. It is rotationally symmetrical about the point (3, 3) (order 2) except for the 'holes'.

That should get you a good start I think

Answer 2

It will look like the graph of y=3(x+2)/(x+3) with holes at x=-2 and x=2. Whenever you would cancel with simplifying, you will get a hole.

Answer 3

Plug in a value for X and plot the point. Plug in a different value for X and plot that point. Use several values for X; positive, negative, and fractional.