How many asymptotes are associated with the graph of h(x) ?..................................................?

Question

f(x) = x2 + 3x + 4.
g(x) = (x2 - 1)(x2 - 4).
h(x) = [f(x)]/[g(x)].



A. none
B. one
C. two
D. three
E. four
F. five

f(x) = x^2 + 3x + 4. g(x) = (x^2 - 1)(x^2 - 4). h(x) = [f(x)]/[g(x)].

and how about
G. None of these

Answer 1

E

Hint: Where is the demoninator (g(x)) zero?

Answer 2

Asymptotes take place the place the linked fee of h(x) methods infinity i.e whilst g(x)=0. From there its trouble-free equation to clean up. (x^2-4)(x^2-9)=0 So x=±2 or x=±3 So in finished there are 4 asymptotes

Answer 3

You get an asymptote when you divide by zero. Since f and g are well-behaved, all we have to look at is h. Now, g is the denominator of h, so we need to know the zeroes of g. These roots are {-2, -1, +1, +2}. For those four values, the denominator goes to zero.

As a check, let's see if we get any 0/0 situations.
f(x) = x^2 + 3x + 4 doedsn't factor, so that's not a problem.

There are four asymptotes, one for each x-value in the set of roots.

Answer 4

I count four but the real way to solve is by looking at the function.

An asymptote is anywhere the function is undefined or its limit is equal to + or - infinity. by looking at your function the graph will have four asmyptotes.

two at the solution of x^2 = 1
and
two at the solution of x^2 = 4

These solutions are 1, -1, 2, -2 so these are your asymptotes.

So the answer is E

Answer 5

e. Anything that will result in division by 0 will be an asymptote.

If you factor the dedominator of h(x), which is g(x) = (x2-1)(x2-4), you get (x+1)(x-1)(x+2)(x-2).
Setting these each equal to 0 we get x={-2,-1,1,2}, meaning that if x is any of these values, we will end up dividing by 0.

So f(x) has 4 aymptotes, which will prevent x from ever being equal to -2,-1,1,2.

Answer 6

E
jogimo has a good explanation.

Answer 7

h. Q asymptotes. now do ur own homework. lol