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 July 31st, 2012, 02:00 AM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Rational function with parameters hey, I have a really difficult question I've been tackling all morning. Appreciate any help you can give me. The slope of the tangent line to the function y=ax/(x^2-bx+ at the point (3,-3) is -1/10. find parameters a and b. Link to word documents so its easier to see the function (u guys gotta teach me how to be able to write equations on this forum) http://s137.photobucket.com/albums/q225 ... lculus.jpg What I've tried to do is, ive put 3 as my x and -3 as my y, and found the equation a=3b-17 which is true. but when I derive this function to find either a or b, i find that b=8.57... and the answer in the book is b=9 and a=10 please help, show me the way to solve this.
 July 31st, 2012, 02:36 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rational function with parameters We are given: $y=\frac{ax}{x^2-bx+8}$ hence: $\frac{dy}{dx}=\frac{$$x^2-bx+8$$(a)-(ax)(2x-b)}{$$x^2-bx+8$$^2}=\frac{a$$8-x^2$$}{$$x^2-bx+8$$^2}$ When $x=3$ we are told the derivative is equal to -1/10, hence: $\frac{a$$8-3^2$$}{$$3^2-b(3)+8$$^2}=-\frac{1}{10}$ $\frac{a}{(17-3b)^2}=\frac{1}{10}$ $10a=(3b-17)^2$ We are also given the point (3,-3) on the curve, so we know: $-3=\frac{a(3)}{3^2-b(3)+8}$ $3b-17=a$ Substituting for a into the first equation yields: $10a=a^2$ $a(a-10)=0$ We discard the root $a=0$ as this does not satisfy the given conditions. Hence we are left with $a=10$ and so: $b=9$
July 31st, 2012, 02:46 AM   #3
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Re: Rational function with parameters

Quote:
 Originally Posted by MarkFL We are given: $y=\frac{ax}{x^2-bx+8}$ hence: $\frac{dy}{dx}=\frac{$$x^2-bx+8$$(a)-(ax)(2x-b)}{$$x^2-bx+8$$^2}=\frac{a$$8-x^2$$}{$$x^2-bx+8$$^2}$
please explain. Ok I know how to derive a rational function, but how did you get from the derivative to the right side of your equation, if you know what I mean.

 July 31st, 2012, 02:51 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rational function with parameters Let's look at the numerator, immediately after applying the quotient rule. I assume you are asking how this was simplified. $$$x^2-bx+8$$(a)-(ax)(2x-b)$ Notice both terms have the common factor a, so let's factor that out: $a$$\(x^2-bx+8$$-x(2x-b)\)$ Now, remove the extraneous parentheses and distribute the $-x$: $a$$x^2-bx+8-2x^2+bx$$$ Collect like terms: $a$$8-x^2$$$
 July 31st, 2012, 02:54 AM #5 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Rational function with parameters ah common factor a that's right. Alright I get it. Thank you!
 July 31st, 2012, 02:57 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rational function with parameters Glad to help, and a belated welcome to the forum!

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