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Find the equation of the normal to the curve y = 3x^2-2x-1 which parellel to the line y = x-3

Mathematics
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find y', set it equal to -1 and solve for x; that's an x coord. on your line. plug into y = 3x^2-2x-1to get a y coord. you have a point and the slope now.. find the line?
1º. Find the derivative of y 2º Put it in some form you like (slope intercept, general...) 3º change it's direction vector to perpendicular 4º find x, for which it's slope is 1

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Other answers:

i got the points, why is the slope 1, it has to be -1 since we take the normal line?
if slope of the tangent = -1, slope of the normal =1.
can u show it in a diagram?
|dw:1347988256856:dw|
i cant picture the problem clearly?
@Algebraic! where the parellel like y=x-3?
wut
can u show the parellel line y =x-3 in the diagram?
|dw:1347988441351:dw|
this was just a general sketch to illustrate the point. this isn't your parabola. but that's the line x-3 in black
why cant we directly get the gradient as m=1 instead of -1
try it and see
im getting a different answer and the answer is wrong
what's your answer?
4y =2x-5
dy/dx= 6x - 2 6x - 2 = 1 x= 1/2 y= -5/4
cn u shw the working
find y', set it equal to -1 and solve for x; that's an x coord. on your line. plug into y = 3x^2-2x-1to get a y coord. you have a point and the slope now.. find the line?
why do we set it equal to -1 instead of 1?
you don't like negative one so you went with one instead? and got the wrong answer?
when the tangent to the parabola has a slope of -1, the normal at that point has a slope of 1
i got it thanks :)

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