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

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.

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

First, let's find the derivative of the function y = x/(x + 5) with respect to x. The derivative will give us the slope of the tangent line at any point on the curve.

Using the quotient rule, we get:

dy/dx = [(x + 5)(1) - x(1)] / (x + 5)^2
= 5 / (x + 5)^2

Now, let's find the slope of the tangent line at the point (2, 7) by substituting x = 2 into the derivative:

m = dy/dx = 5 / (2 + 5)^2
= 5 / (7)^2
= 5 / 49

We now have the slope of the tangent line that passes through (2, 7). Let's denote it as m1.

Next, using the point-slope form of a linear equation, we can write the equation of the tangent line passing through (2, 7):

y - y1 = m1(x - x1)

Substituting the values, we have:

y - 7 = (5 / 49)(x - 2)

Expanding the equation:

y - 7 = (5 / 49)x - 10 / 49

To find the second tangent line, we'll consider a slightly different approach. We'll find the slope of the tangent line at the point (2, 7), denoting it as m2, and then use the point-slope form of a linear equation again.

To find the slope, we can choose another nearby point on the curve, say (2 + h, (2 + h) / ((2 + h) + 5)), where h is a small value.

Using the derivative formula from above, we can find the derivative at x = 2 + h by substituting x = 2 + h:

m2 = dy/dx = 5 / ((2 + h) + 5)^2

Expanding the denominator:

m2 = 5 / (h^2 + 9h + 14)

Now, let's find the equation of the tangent line passing through (2, 7) using the point-slope form:

y - 7 = m2(x - 2)

Substituting the slope m2 and expanding the equation gives:

y - 7 = (5 / (h^2 + 9h + 14))(x - 2)

This is the equation of the second tangent line. Note that as h approaches 0, the equation will become the tangent line passing through (2, 7).

To summarize, the equations of the tangent lines to the curve y = x/(x + 5) that pass through the point (2, 7) are:

y - 7 = (5 / 49)(x - 2)

y - 7 = (5 / (h^2 + 9h + 14))(x - 2) (as h approaches 0)