Answer:
it stays the same because there is no simplifying it and the re are no like terms
Answer:
77
Step-by-step explanation:
If you have any questions about the way I solved,don't hesitate to ask
16 is the answer to this question.
Answer:

Step-by-step explanation:
<u>Composite Function
</u>
Given f(x) and h(x) real functions, the composite function named
is defined as:

It can be found by substituting h into f.
We are given the functions:


It's required to find 
As defined above:

Thus:

Find h(2):

h(2)=0
Now:


Ok so we need to set the bottom to 0 to find the vertical asympyotes. This becomes x^2 - 4 = 0. Since we're talking about asymptotes, i'll assume you can solve basic equations. Solving for x and you get x = ±2. This means the vertical asymptotes are at ±2. To solve for horizontal asymptotes you take the limit as x goes to ±∞. Either way you end up with ±∞/∞. Now this isn't 1 because they grow at different rates. You differentiate both the top and bottom(L'hopital) and you get 6x/2x which becomes 3. This means the horizontal asymptote is at y = 3.