This is a question on my math homework, but I can't make any sense of it. Here it is: find the equation of the normal to the curve with equation y=(x^3)+1. Is there a typo or am I just missing something?

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This is a question on my math homework, but I can't make any sense of it. Here it is: find the equation of the normal to the curve with equation y=(x^3)+1. Is there a typo or am I just missing something?

Mathematics
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The normal is a line perpendicular to the tangent line
Could you explain how to do this problem? It has been a while since I have done this kind of problem.
for what point are you trying to find the normal to?

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Other answers:

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)
k... now take the derivative of you function.... plug in the values (1,2) this will give you the slope, Let me know what you get
find f'(x) and evaluate it at x=1 then take the negative reciprocal of that this gives you the slope of the normal line
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
since the normal is perpendicular to the tangent, take the reciprocal of the slope of the tangent to get the slope of 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?
wait the slope is -1/3 bc perpendicular slope have opposite signs and the point slope formula is \[y-y_1=m(x-x_1)\]
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?
no... so you have \[y-2=-\frac{1}{3}(x-1)\] Expand this and you get \[y-2=-\frac{1}{3}x+\frac{1}{3}\] now combine like terms \[y=-\frac{1}{3}x+\frac{1}{3}+2\] \[y=-\frac{1}{3}x+\frac{7}{3}\] which is in the form of y=mx+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

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