Finding Equation of Tangent Line to the Curve

[Dazed]

1. Find the equation of the tangent line to the curve
y= f(x)= (3+x)/(x-2) at the points where x = 3, 7, -1, and -3.


2. Graph y=f(x)= (3+x)/(x-2) carefully giving the domain, range, all asymptotes, and atleast 8 points with both coordinates.

 

[soroban]

Hello, Dazed!

I will assume you know the Quotient Rule for derivatives
and you know the Point-Slope Formula for the equation of a line.
[If you don't, review them ... quickly!]

1. Find the equation of the tangent line to the curve
y= f(x)= (x+3)/(x-2)
at the points where x = 3, 7, -1, and -3.

When x = 3: .y .= .f(3) .= .(3 + 3)/(3 - 2) .= .6.
. . We have a point: .(3,6)

The derivative gives us the <u>slope</u> of the tangent line.

. . . . . . . . . . . (x - 2)·1 - (x + 3)·1 . . . . . . -5
. . . f '(x) . = . ------------------------- . = . ----------
. . . . . . . . . . . . . . . (x - 2)² . . . . . . . . . .(x - 2)²

When x = 3: .f '(3) .= .-5/(3-2)² .= .-5


The line through (3,6) with slope -5 is:

. . . y - 6 .= .-5(x - 3) . . ---> . . y .= .-5x + 21