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The normal is a line perpendicular to the tangent line

for what point are you trying to find the normal to?

Also if the way I have put in the equation is confusing it should be: \[y=x^3+1\]

thats what i was fixing to ask lol

(1,2)

for the derivative I got \[y=3x^2\]

and now I put in 2 for y and 1 for x, such that \[2=3(1)^2\]

correct?

no..... sorry just 1 for x, bc y' is the slope

so then y=3

and to get x i plug in 3 for y in the derivative?

yeah thats your slop for the tanget, now you have to find the slope for the normal

So if the sope of the tangent is 3, then the slope of the normal is 1/3, correct?

close... -1/3

If the reciprocal is the number written as a fraction then flipped why did the sign change?

sorry it is 1/3, I'm being retarded

and then thats it? the answer is 1/3?

now use the point slope formula to get the equation of the normal

refresh my memory, what is the point slope formula? Or can i just use y=mx+b?

perpendicular lines are negative reciprocals

ok. y1 and x1 are from the points given at the start, right?

yes

and m=\[-1/3\]

yes

so plugging everything in gets\[3-2=-1/3(1-1)\]

wich solves to 1=0

no.... you are looking y, the values you have are y1, x1, and m

y1=2, x1=1 and m=-1/3

so then it should be \[y-2=-1/3(x-1)\]

yeah... now put it in terms of y=mx+b

\[2=-1/3(1)+b\]

and solve for b?

could you also distribute the negative and move the 1/3 from \[y-2=-1/3(x-1)\]
to give 3y-6=-x+1

then move -x+1 over to give x+3y-7=0

yes you could

ok thank you for the help. yo just got a new fan!

also now i'll use this site for help with math if the teacher is unavailable. Thanks again!

Cheers , hope I didin't confuse you

no you did fine. Have to go now bye