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?

- anonymous

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- anonymous

The normal is a line perpendicular to the tangent line

- anonymous

Could you explain how to do this problem? It has been a while since I have done this kind of problem.

- anonymous

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

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## More answers

- anonymous

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

- myininaya

thats what i was fixing to ask lol

- anonymous

(1,2)

- anonymous

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

- myininaya

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

- anonymous

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

- anonymous

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

- anonymous

correct?

- anonymous

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

- anonymous

so then y=3

- anonymous

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

- anonymous

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

- anonymous

since the normal is perpendicular to the tangent, take the reciprocal of the slope of the tangent to get the slope of the normal

- anonymous

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

- anonymous

close... -1/3

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

wait the slope is -1/3 bc perpendicular slope have opposite signs and the point slope formula is
\[y-y_1=m(x-x_1)\]

- anonymous

perpendicular lines are negative reciprocals

- anonymous

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

- anonymous

yes

- anonymous

and m=\[-1/3\]

- anonymous

yes

- anonymous

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

- anonymous

wich solves to 1=0

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

and solve for b?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes you could

- anonymous

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

- anonymous

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

- anonymous

Cheers , hope I didin't confuse you

- anonymous

no you did fine. Have to go now bye

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