Question

How do you find the equation of the line tangent to the graph of `Y= (X-3) / (X-4)` at (5,2)?

Answer 1

`y-2 = -1(x-5)` or `y = -x+7`

Explanation:

The challenge, of course, is to find the slope of the tangent line. Once we know that the slope of the tangent line at that point is `-1`, writing an equation for the line is straightforward.

There are shortcut methods, simplified rules and properties for finding slopes of tangent lines.

If you haven't learned them yet, then you are probably working with one of the definitions for the slope of the line tangent to the graph of `y=f(x)` at the point `(a, f(a))` is (either of these;)

`lim_(xrarra)(f(x)-f(a))/(x-a)" "`or `" " lim_(hrarr0)(f(a+h)-f(a))/h`

(If one of these limit exist, then both do and they are equal.)

For this problem we have `f(x) = (x-3)/(x-4)` and `a = 5`

So we want

`lim_(xrarr5)(f(x)-f(5))/(x-5)" "`or `" " lim_(hrarr0)(f(5+h)-f(5))/h`

If we try to evaluate the limit by substitution, we get the indeterminate form `0/0`, so we need to do some algebra. We were given (and can easily verify) that `f(5) = 2`)

`lim_(xrarr5)(f(x)-f(5))/(x-5) = lim_(xrarr5)(((x-3)/(x-4))-2/1)/(x-5)` `" "`form is `0/0`

`= lim_(xrarr5)(((x-3)-2(x-4))/(x-4))/(x-5)` `" "`form is `0/0`

`= lim_(xrarr5)(-x+5)/((x-4)(x-5))` `" "`form is `0/0`

`= lim_(xrarr5)(-1(x-5))/((x-4)(x-5))` `" "`form is `0/0`

`= lim_(xrarr5)(-1)/(x-4) = -1/1 = -1`

Now that we have slope `m=-1` and point `(5,2)` we can get the equation of the line.

`y-2 = -1(x-5)`

`y-2 = -x+5`

`y = -x+7`

Using the quotient rule and other rules:

`y = (x-3)/(x-4)`

`y' = (1(x-4)-(x-3)(1))/(x-4)^2 = (-4+3)/(x-4)^2 = (-1)/(x-4)^2`

At `x=5`, we get `y'|_(x=5) = (-1)/(1)^2 = -1`

Answer 2

The answer is `y-2=-(x-5)` in point-slope form or `y=-x+7` in regular xy-form.

Explanation:

To find the equation of the tangent line to any graph, these rules must apply:

1. Find the derivative of the function for any `x`. Remember the idea of secant lines coming together to one point (the instantaneous rate of change or the tangent line).
2. Given the point, insert the `x` to find the derivative of the function at that point (which will give you the slope of the tangent line).
3. Use point-slope form to find the equation of the tangent line.

First off, the equation `y=(x-3)/(x-4)` is difficult to take the derivative directly.
We can simplify it to when we can use the Power Rule and the Chain Rule to solve for the derived function of `x`, using the idea of clever numbers:

`y=(x-3)/(x-4)=(x-4+1)/(x-4)=(x-4)/(x-4)+1/(x-4)=1+1/(x-4)`.

So `y=1+(x-4)^-1`, which means that its derivative is
`dy/dx=-(x-4)^-2` for any `x`.

Now we need the to include the point `(5,2)` and find out the slope of the tangent line at `x=5`:

`dy/dx|_(x=5)=-(5-4)^-2=-1`

Finally we can use point-slope form to find the equation of the tangent line:

`y-y_1=m(x-x_1)`, where `m` is slope, `y_1` is the y-point, and `x_1` is the x-point.

For the function `y=(x-3)/(x-4)`, the tangent line at `(5,2)` is
`y-2=-(x-5)`,
`y=-(x-5)+2`,
or `y=-x+7`.

You can always check to see if it works with a graphing calculator, but the key thing is to understand why you made a mistake for better confidence.