Question

How do you find the equation of the tangent line to the curve `y= (x-3) / (x-4)` at (5,2)?

Answer 1

`y=-x+7`
Refer below for explanation.

Explanation:

Step 1: Take the Derivative
This requires use of the quotient rule:
`d/dx(u/v)=(u'v-uv')/v^2`
Where `u` and `v` are functions of `x` and`!=0`.

In our case,
`u=x-3->u'=1`
`v=x-4->v'=1`

Therefore,
`y'=((x-3)'(x-4)-(x-3)(x-4)')/(x-4)^2`
`y'=(1(x-4)-(x-3)(1))/(x-4)^2`
`y'=(x-4-x+3)/(x-4)^2`
`y'=-1/(x-4)^2`
Note that this will always be negative, which means `y` is always decreasing, which further implies that the slope of the tangent line will be negative.

We can now proceed to step 2.

Step 2: Evaluate the Derivative
We are being asked the find a tangent line, and one component of a tangent line is the slope. We find the slope by evaluating the derivative at a point (in this case, `x=5`):
`y'=-1/(x-4)^2`
`y'(5)=-1/((5)-4)^2=-1/(1)^2=-1`
This tells us the slope at `(5,2)` is `-1`.

Step 3: Find the Equation
Tangent lines are of the form `y=mx+b`, where `x` and `y` are points on the line, `m` is the slope, and `b` is the `y`-intercept. We have everything except `b`, so we need to find it. Luckily, we have a point `(5,2)` and a slope (`-1`), which we can use to determine `b`:
`y=mx+b`
`2=(-1)(5)+b->2=b-5->b=7`

The equation of the tangent line is thus `y=-x+7`.