Equation of the normal line

[cherica123]

Find the equation of the normal line to the graph of y=-3ln|x| at x=-3

What I did:

took the derivative to get y=-3/x

plugged in -3 for x to get y=1

and then from here im not sure what to do...if that work is correct thus far. Could someone guide me the rest of the way? A problem like this will be on our test so I have to understand it. Thanks!

 

[galactus]

The normal is perpendicular to the tangent line at that point.

So if your slope at x=-3 is 1, your normal slope is -1.

Use y=mx+b to find the normal line equation.

You know a point it pases thorugh and the slope.

 

[cherica123]

of so using y=mx+b....can i plug in -3ln|x| for y? and x=-3

-3ln|3|=-3(-1)+b

-3.30-3+b

-6.30=b

so y=-x-6.3


is that correct?

 

[galactus]

Yep, looks right. Good work.

Except maybe write your answer as:

\(\displaystyle \L\\y=-x-3(ln(3)+1)\)

instead of in decimal form. Looks better and is more exact.

 

[cherica123]

awesome, thanks galactus! and good point, we wont have a calculator on our test, so duh i cant use the decimal form.