# Thread: family of functions- verticle asymptotes

1. ## family of functions- verticle asymptotes

im stuck on a question!

a family of functions is given by:
r(x)=1/a+(x-b)^2

a) for what vales of a and b does the graph of r have a verticle asymptote? where are the verticle asymptotes in this case?
b) find values of a and b so that the function r has a local maximum at the point (3,5)

any help would be greatly appreciated
thanks

2. Hello,

I'm not sure what you mean:

$r(x)=\frac1a + (x-b)^2$

or

$r(x)=\frac{1}{a + (x-b)^2}$

3. the second one!

4. Have you tried solving a + (x - b)^2 = 0? I think that would pin it down for you.

5. Originally Posted by Paige05
im stuck on a question!

a family of functions is given by:
r(x)=1/(a+(x-b)^2)

a) for what vales of a and b does the graph of r have a verticle asymptote? where are the verticle asymptotes in this case?
b) find values of a and b so that the function r has a local maximum at the point (3,5)
A function has vertiacal asymptotes if the denominator equals zero:

$a+(x-b)2=0~\iff~(x-b)^2=-a~ \iff~x=b+\sqrt{-a}~ \vee~x=b-\sqrt{-a}$

That means:

you get no vertical asymptote if a > 0

you get exactly one vertical asymptote if a = 0 then the equation of the asymptote is : x = b

you get 2 vertical asymptotes if a < 0. The equations of the asymptotes are: $x=b+\sqrt{-a}~ \vee~x=b-\sqrt{-a}$

to b):

$r(x)=(a+(x-b)^2)^{-1}$ Use chain rule to derivate:

$r'(x)=(-1)(a+(x-b)^2)^{-2} \cdot (2(x-b))$ Rearrange:

$r'(x)=-\frac{2(x-b)}{(a+(x-b)^2)^{2}}$

$r'(x) = 0 ~\implies~2(x-b)=0~\iff~x=b$

From your problem you know that now b = 3. Calculate

$r(3)=\frac{1}{a+(3-3)^2} = 5 ~\implies~a = \frac15$

6. ahhhhh.... i see!
thanks so much