Question

Consider the following curve.
y = (x+4)/(x-4).
Find an equation of the tangent line to the curve at point (3, 7).

Answer 1

To find the equation of the tangent line, we need to find the slope of the curve at point (3, 7).

The slope of a curve at a given point can be found by taking the derivative of the curve and evaluating it at that point.

Let's find the derivative of the curve y = (x+4)/(x-4).

To simplify the process, let's rewrite the curve as y = (1 + 8/(x - 4)).

Now, we can use the quotient rule to find the derivative of the curve.

Let u = 1 and v = 8/(x - 4).

The derivative of the curve y = (1 + 8/(x - 4)) can be found using the quotient rule:

dy/dx = (v*u' - u*v') / v^2

where u' and v' represent the derivatives of u and v with respect to x.

u' = 0 (since the derivative of a constant is 0)

v' = -8/(x - 4)^2 (using the power rule for derivatives)

Substituting these values into the quotient rule, we have:

dy/dx = (v*0 - u*(-8/(x-4)^2)) / v^2
= 8/(x-4)^2

Now, let's evaluate the derivative at point (3,7):

dy/dx = 8/(3-4)^2
= 8/(-1)
= -8

So, the slope of the curve at point (3,7) is -8.

Now, we can use the point-slope form of a line to find the equation of the tangent line:

y - y1 = m(x - x1)

where m represents the slope of the line and (x1, y1) represents the coordinates of the point.

Plugging in the values, we have:

y - 7 = -8(x - 3)

Simplifying the equation:

y - 7 = -8x + 24

y = -8x + 31

Therefore, the equation of the tangent line to the curve at point (3, 7) is y = -8x + 31.