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Mathematics 6 Online
OpenStudy (anonymous):

Find an equation of the tangent line to the given curve at this point 1. y= 1 + 2x - x^3 (1,2)

OpenStudy (anonymous):

find y' it tells you the slope of the function for any x. the slope of the function at x=1 will be y ' (1) so that will also be the slope of the tangent line at that point. you now have 'm' in ' y=mx+b' use the point x=1 and y=2 to find the value of 'b'

OpenStudy (anonymous):

i still dont understand.. thats not what my teacher taught. we need to find the slope some how byt using f(x)-f(a)/x-a

OpenStudy (anonymous):

ok, so you haven't learned rules of differentiation yet?

OpenStudy (anonymous):

no not yet

OpenStudy (anonymous):

\[\lim_{a \rightarrow0} \frac{1 + 2x - x^3 -(1 + 2a - a^3)}{ x-a }\]

OpenStudy (anonymous):

\[\lim_{a \rightarrow0} \frac{ 2x - x^3 - 2a +a^3}{ x-a }\]

OpenStudy (anonymous):

\[\lim_{a \rightarrow0} \frac{ -(x^3 -a^3) + 2x- 2a }{ x-a }\]

OpenStudy (anonymous):

use the difference of cubes formula on (x^3 -a^3) \[a^3 – b^3 = (a – b)(a^2 + ab + b^2)\]

OpenStudy (anonymous):

\[\lim_{a \rightarrow x} \frac{ - (x – a)(x^2 + ax +a^2) +2(x-a)) }{(x-a) }\]

OpenStudy (anonymous):

I had a->0 in those earlier limits, it's supposed to be a->x

OpenStudy (anonymous):

need me to retype them or is it ok?

OpenStudy (anonymous):

no your fine i understand so far

OpenStudy (anonymous):

that's about it really... (x-a) is in both terms in the numerator and also in the denominator. cancel it. then take the limit...

OpenStudy (anonymous):

so you get (a^2 + ax + x^2 ) + 2

OpenStudy (anonymous):

yep:)

OpenStudy (anonymous):

but that isnt an equation

OpenStudy (anonymous):

take the limit.

OpenStudy (anonymous):

what do you mean take the limit the limit for that

OpenStudy (anonymous):

we're finding the derivative here, remember? that's step one. once you have the derivative you have the slope of the function for any x... find y' it tells you the slope of the function for any x. the slope of the function at x=1 will be y ' (1) so that will also be the slope of the tangent line at that point. you now have 'm' in ' y=mx+b' use the point x=1 and y=2 to find the value of 'b'

OpenStudy (anonymous):

it never says x is one in my problem though

OpenStudy (anonymous):

"at this point (1,2)"

OpenStudy (anonymous):

so why did i just do all that work if i can just plug in 1 in the first place

OpenStudy (anonymous):

go for it.

OpenStudy (anonymous):

but i just really dont understand what the hell im doing your telling me i can do so many different things. like wtf

OpenStudy (anonymous):

why don't you show me how you'd do the problem.

OpenStudy (anonymous):

i dont know thats the problem thats why i came on here

OpenStudy (anonymous):

And I just showed you.

OpenStudy (anonymous):

but you don't seem to like it; you have an easier way. so show me.

OpenStudy (anonymous):

im asking you i dont know what to do! none of this makes sense to me, i understnand it all in class and then when it comes to homework i sit for hours and not know what im doing like its to the point where im cying

OpenStudy (anonymous):

do you understand that the tangent line at a point has the same slope as the function at that point?

OpenStudy (anonymous):

nevermind forget thank you. ill just ask my teacher tomorrow.

OpenStudy (anonymous):

k.

OpenStudy (anonymous):

no need to be a fluttering feather i said thank you.

OpenStudy (anonymous):

blocked.

OpenStudy (anonymous):

lol bye (:

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