Equation of a tangent using derivatives

S

Shaughnessy

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Mar 13, 2006
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I was just wondering how i would go about solving this problem:

Find the equation of the tangents from the origin to the graph of

y= (x+9)/(X+8)

I took the derivative which would be the slope right?? using the quotient rule and got a -1/(x+8)^2 but am unsure of where to go from here. I was able to do the question before when it was asked to find the tangent to the graph that intersects the coordinate axis, but I have no clue what to do here. Would the y intercepts be 9/8?? and would that help me solve the question in any way?

Any help would definately be appreciated!
Thanks,
Shaughnessy
 
Notice first that (0,0) is NOT on your curve. It is likely that x=0 has very little to do with the slope of the tangent lines you seek. The lines you seek pass through (0,0), but are tanget to the curve for some other value of x. You have the right derivative. You do NOT know what value of x to use. Use an arbitrary value and solve.

Your lines look like this, using the point-slope form: y-0 = m*(x-0) or just y = m*x.

What is m? It is the slope of the tangent line at SOME point. Let's call the point (a,b). The derivative is -1/(x+8)^2, evaluated at x=a, gives -1/(a+8)^2. You also know that b = (a+9)/(a+8), since (a,b) is on the curve.

y = m*x becomes (a+9)/(a+8) = (-1/(a+8)^2)*a

Solve for 'a', then figure out what it all means.
 
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