InterserveVB said:
I have a slight problem. I am trying to figure out how to find the tangent lines to the curve y = x/(x+2) that pass through (8, 9). The answer is y = 1.12x + 0.06, but I am not sure how to get that.
y' = [(x+2)*1 - x*(1)]/(x+2)<sup>2</sup> = 2/(x+2)<sup>2</sup>
You have two points, one you know precisely, and one for which you've these two hints.
y = x/(x+2)
y' = 2/(x+2)<sup>2</sup>
Let's say the point we're looking for is (a,b).
b = a/(a+2)
m = 2/(a+2)<sup>2</sup> -- The slope of the tangent line
And we have y-b = [2/(a+2)<sup>2</sup>](x-a) = y-[a/(a+2)]
This line must pass through (8,9)
9-[a/(a+2)] = [2/(a+2)<sup>2</sup>](8-a)
I get a = ¼(sqrt(41)-9) or a = ¼(-sqrt(41)-9), both of which appear to lead to an appropriate solution. In addition to the solution you have been given, you may wish to investigate y = 0.584*x + 4.329
Speaking of that solution you have been given, it doesn't look quite right. Perhaps y = 1.096*x + 0.231 would be better.
