SOLUTION: a). Find the function f such that f'(x) = 2sin(x) + sec^2(x) and f(0) = 3 b) Find the horizontal asymptote(s) of the graph of: y = (2x^2+5)/(3-x^2) I got: y=-2 because for

Algebra ->  Finance -> SOLUTION: a). Find the function f such that f'(x) = 2sin(x) + sec^2(x) and f(0) = 3 b) Find the horizontal asymptote(s) of the graph of: y = (2x^2+5)/(3-x^2) I got: y=-2 because for       Log On


   

Question 999328: a). Find the function f such that f'(x) = 2sin(x) + sec^2(x) and f(0) = 3
b) Find the horizontal asymptote(s) of the graph of: y = (2x^2+5)/(3-x^2)
I got: y=-2 because for horizontal you check the exponents and they are the same here so -2 would make the most sense, I think.
a I have no idea how to solve
Please explain
Thank you
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Integrating the function gives you.
df%2Fdx=2sin%28x%29%2Bsec%5E2%28x%29
f=-2cos%28x%29%2Btan%28x%29%2BC
f%280%29=-2%2A1%2B0%2BC=3
-2%2BC=3
C=5
f%28x%29=tan%28x%29-2cos%28x%29%2B5
.
.
.
Yes, correct.
y=-2
.
.
.
.