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

##### 2 Answers

#### 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

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

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

For this problem we have

So we want

If we try to evaluate the limit by substitution, we get the indeterminate form

# = 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

Using the quotient rule and other rules:

At

The answer is

#### Explanation:

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

- 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). - 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*). - Use
**point-slope form**to find the equation of the tangent line.

First off, the equation

We can simplify it to when we can use the **Power Rule** and the **Chain Rule** to solve for the derived function of *clever numbers*:

So

*any*

Now we need the to include the point

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

For the function

or

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.