please help with calc?

Question

let f be the function that is given by f(x) = (ax+b) / (x^2 - c) and has the following properties:
1) the graph of f is symmetric to the y axis
2)lim as x approaches 2 from the right = + infinity
3)f ' (1)= -2

What are the values of a, b, and c?
What are the vertical and horizontal asymptotes?


thank you sosososososososososo much

Answer 1

a=0
c=4
b=9

From 1,
(a*x+b) / (x^2-c) = (a*(-x)+b) / ((-x)^2-c) = (-a*x+b) / (x^2-c)
which implies a*x = -a*x or a=0, so
y = b/(x^2-c)

From 2, by inspection,
c=4 and b>0, so
y = b/(x^2-4)

From 3,
y' = -2*b*x / (x^2-4)^2 so
-2 = y'(1) = -2*b*1 / (1-4)^2 = -2*b/9 which implies b=9.

You can figure out the asymptote part given,
y = 9 / (x^2 - 4)
Graph function at link below,

Answer 2

f(x) = 9/(x^2-4)

a=0, b=9, c=4

vertical asymtotes x=2, x= -2

horizontal asymptote y=0 (as x---> +or- inf, f(x)---> 0)

1) a=0..If f(x) is symmetric about the y-axis then f(x) = f(-x)
which implies -ax+b = ax+b so a =0

2) c=4 since x^2 -4 =(x+2)(x-2) so if x--> 2 from the right x-2 will approach 0 but will be +ve so the denominator approaches 0 and is +ve so the function ----> + inf

3) We have determined f(x) = b /(x^2-4)

f'(x) = -2xb / (x^2-4)^2
we are given f'(1) = -2 so we can determine b:

-2 = -2b/(1-4)^2 implies b = (1-4)^2 = 9

Answer 3

1) symterical in the y-axis means : f(x) = f(-x) from this observation it follows immediatly that b = 0 or a = 0

2) thus c = 4

3) the rest you can do yourself. ( i dont like to differentiate in a textbox)