Equation of a normal line...

[gaminlegend38]

Alright, seems simple enough, but apparently I'm getting the wrong answer.

An equation of the normal line to the graph of f(x)=x/(2x-3) at the point where x=1 is...

A. 3x+y=4
B. 3x+y=2
C. x-3y=-2
D. x-3y=4 (supposedly right answer)
E. x+3y=2

So I quotient rule to find the derivative:
((2x-3)(1)-(x)(2))/(2x-3)^2

(2x-3-2x)/(2x-3)^2

-3/(2x-3)^2

Plug in the one and get 3. So the normal would then be -1/3. The original point would be (1,-1).

y+1=-1/3(x-1)
y+1=-1/3x + 1/3
1=-1/3x - y +1/3
2/3=-1/3x-y

Obviously not one of the answers. Where did I go wrong?

 

[skeeter]

Alright, seems simple enough, but apparently I'm getting the wrong answer.

An equation of the normal line to the graph of f(x)=x/(2x-3) at the point where x=1 is...

A. 3x+y=4
B. 3x+y=2
C. x-3y=-2
D. x-3y=4 (supposedly right answer)
E. x+3y=2

So I quotient rule to find the derivative:
((2x-3)(1)-(x)(2))/(2x-3)^2

(2x-3-2x)/(2x-3)^2

-3/(2x-3)^2

Plug in the one and get 3. check this again

So the normal would then be -1/3. The original point would be (1,-1).

y+1=-1/3(x-1)
y+1=-1/3x + 1/3
1=-1/3x - y +1/3
2/3=-1/3x-y

Obviously not one of the answers. Where did I go wrong?

 

[Dr. Flim-Flam]

f(x) =x/(2x-3), f '(x) = -3/(2x-3)^2

Let g(x) = slope of normal line, then g(x) = (2x-3)^2/3 g(1) = 1/3, f(1) =-1

ergo y-(-1) = (1/3)(x-1), y = (x-4)/3 or 3y =x-4.

Note: the slope of the normal line g(x) = the negative reciprocal of the slope of f(x).

g(x) *f '(x) = -1