find equations of the tangent lines to the curve y=x/(x+5) that pass through the point (2,7).

Answer 1

y = x/(x+5)
y' = [(x+5) - x]/(x+5)^2 = 5/(x+5)^2

So, we want lines through passing through (x,y) and (2,7) with slope y'

So, for any x, we have

(y-7)/(x-2) = 5/(x+5)^2

x/(x+5) - 7 = 5(x-2)/(x+5)^2

x(x+5) - 7(x+5)^2 = 5(x-2)

x^2 + 5x - 7x^2 - 70x - 175 = 5x - 10

6x^2 + 70x + 165 = 0

has roots x = -8.388 and -3.278

y(-8.388) = 2.475
y(-3.278) = -1.903

The line through (-8.388,2.475) and (2,7) has slope 4.525/10.388 = 0.436

5/-3.388^2 = 0.436

So, now you have a point and a slope, and you can write the equation for that line.

Do the same for the other value of x.