Question

find the equation of the normal line to the curve y=ln(x^2-3) at x=2 (step by step)
WILL FAN AND MEDAL!

Answer 1

first find the tangent, and also the value of f(2)

Answer 2

take the derivative .... y' defines the slope of the tangent at a given point

Answer 3

would the derivative be ln(2x)??? @amistre64

Answer 4

the derivative of ln(u) = u'/u

let u = x^2 -3, so u' = 2x

Answer 5

y' = 2x/(x^2-3) .... and at x=2 this is?

Answer 6

i plug the 2 into the x's right? if so it would be -2
@amistre64

Answer 7

x=2

2(2)/(2^2-3) = 4/(4-3) = 4

Answer 8

not real sure how youd get -2 .... but thats beside the point :)

Answer 9

yea......i dont know either but thanks!! @amistre64

Answer 10

the important thing is knowing the slope of the tangent line allows us to get the slope of the normal line by perping it ...

tangent and normal are 90 degrees to each other meaning
if the tangent is m, the normal is -1/m

Answer 11

now tell me what you know about contructing a line using a point and a slope

Answer 12

use rise/run (slope) when the points are plotted on the graph to make a line

Answer 13

@amistre64

Answer 14

sites lagging ...

lets go this route since you gave it a shot

given slope m, and a point (a,b)

the equation of a line in point slope form is constructed as:

y-b = m(x-a)

Answer 15

in terms of functions and their derivatives ..

our tangent line is:
\[y-f(a)=f'(a)(x-a)\]

our normal line is:
\[y-f(a)=-\frac{1}{f'(a)}(x-a)\]

Answer 16

what is f(a) and f'(a) im confused now??? @amistre64