asymptotes

**mer1988** · October 18th 2007, 08:08 PM

A family of functions is given by

Find values of a and b so that the function r has a local minimum at the point (5 , 9)

Find the vertical asymptotes of r with the values of a and b found in part (a).

ok can someone just help me start this one...it would help alot, thanks

mer

**earboth** · October 18th 2007, 08:38 PM

Originally Posted by mer1988

A family of functions is given by

Find values of a and b so that the function r has a local minimum at the point (5 , 9)

Find the vertical asymptotes of r with the values of a and b found in part (a).

ok can someone just help me start this one...it would help alot, thanks

mer

Hello,

1. I can't open the files which probably show what you have done so far

2. Calculate the derivative. Use chain rule:

$r'(x) = (-1)(a-(x-b)^2)^{-2}) \cdot 2(x-b) = -\frac{2(x-b)}{\left(a-(x-b)^2\right)^{2}}$

3. You already know that r'(5) = 0. A fraction equals zero if the numerator is zero (and the denominator is not zero, of course):

$2(5-b) = 0~\iff~b = 5$

4. You already know that the point M(5, 9) is placed on the graph of the function:

$r(x) = \frac{1}{a-(x-5)^2}~\implies~r(5) = \frac{1}{a-(5-5)^2}=\frac1a=9~\implies~a=\frac19$

Thus the equation of the function is: $r(x) = \frac{1}{\frac19 - (x-5)^2}$

5. To find the vertical asymptotes you have to calculate the zeros of the denominator:

$\frac19-(x-5)^2=0$ Solve for x. I've got $x=\frac{14}{3}~\vee~x=\frac{16}{3}$

6. I've attached a diagram of the function.

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