How do you graph and find the discontinuities of #H(x) = [(x - 1) (x + 2) (x - 3)]/(x(x - 4)^2#?

Question

Eli Assouline · Accepted Answer

First we find every assymptote and root, and then start making a table of values. The discontinuities are the vertical assymptotes.

First thing we do, is look for any vertical assymptotes. Those are the values the function can never take, usually because it'll cause a math error (division by zero, logarithm of a number that's zero or negative, negative number in an even root, etc.)

In this case, the only things that can make things go wrong is the denominator. Just make the inequality, and solve it.

#x(x-4)^2 != 0#

Since it's a product, either can't be zero, so
#x != 0#
#(x-4)^2 != 0 rarr (x-4) != 0 rarr x!= 4#

We have two vertical assymptotes, and they are #x = 0# and #x=4#. Make note of it on your graph. So you won't accidentally cross it or something when plotting it later.

The horizontal or slant assymptotes are basically extremes. What happen when the value of x get infinitely big. We could just keep plugging bigger and bigger numbers until we see a pattern, or, we can pretend #x# is a number #k# that is as big as we need, and see if we can find a general value for the assymptotes.

#H(k)=((k−1)(k+2)(k−3))/(k(k−4)^2)#

Since #k# is as big as we want it to be - and we want it to be as big as possible - we can safely say that #k-a~=k# where #a# is a constant value. Basically, #k# is so big that it's always orders of magnitude above the biggest constant on the function, so we can just ignore substractions or additions.

#H(k) ~= (k*k*k)/(k*k^2)=k^3/k^3=1#

Basically, when #|x|# becomes really, really big, #H(x)# has a tendency to go to 1. For all we know, it never becomes 1, or it becomes 1 for a small value of #x# and then it just above 1, it doesn't really matter. The important thing is for big values of x, the function stay close to 1.

So we have a horiontal assymptote (since it's a horizontal line), #y=1#, make note of it on your graph.

Now, since we have our suggestion's guidelines, all that's left is actually picking values and then connecting the dots. We know this function has 3 roots, that is, three points of the form #(x,0)#.

They're #x=1#,#x=-2# and #x=3#. We know this because the function's already conveniently factored. So you can make note of these on your graph too. In the end the graph should look something like this (the middle part will depend on how big the spacing is, but the general outline should be similar):

Abigail Allen · Answer

undefined To graph and find the discontinuities of the function H(x) = [(x - 1) (x + 2) (x - 3)]/(x(x - 4)^2, follow these steps:

1. Determine the x-intercepts by setting the numerator equal to zero and solving for x. In this case, set (x - 1) (x + 2) (x - 3) = 0 and solve for x. The x-intercepts are x = 1, x = -2, and x = 3.

2. Determine the vertical asymptotes by setting the denominator equal to zero and solving for x. In this case, set x(x - 4)^2 = 0 and solve for x. The vertical asymptote is x = 0.

3. Determine the horizontal asymptote by comparing the degrees of the numerator and denominator. Since the degree of the numerator is 3 and the degree of the denominator is 3, divide the leading coefficients of both. The horizontal asymptote is y = 1.

4. Determine the behavior of the function near the vertical asymptote and x-intercepts. Use the sign chart to determine the behavior of the function in each interval.

5. Plot the x-intercepts, vertical asymptote, and horizontal asymptote on the graph.

6. Sketch the graph by connecting the points and following the behavior determined in step 4.

7. To find the discontinuities, look for any values of x that make the function undefined. In this case, x = 0 and x = 4 are discontinuities since they make the denominator zero.

This is how you graph and find the discontinuities of the function H(x) = [(x - 1) (x + 2) (x - 3)]/(x(x - 4)^2.