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[H] Calculus

Blogs > conCentrate9
Post a Reply
conCentrate9
Profile Blog Joined December 2007
United States438 Posts
October 16 2008 23:59 GMT
#1
Okay heres a little problem with derivatives and tangent lines that I need a bit of help on.

For the parabola y = x^2 + x find the equations of both tangent lines that pass through the point (2, -3).

I started by finding the derivative of y as y' = 2x+1 and finding y'(2)=5, but using that slope and the point (2, -3) I get for my tangent line y = 5x - 13, which isn't even close. I just need direction I guess. Thanks.

**
paper
Profile Blog Joined September 2004
13196 Posts
Last Edited: 2008-10-17 00:03:46
October 17 2008 00:03 GMT
#2
so you have the slope of the line [5] and the point (2, -3)

just make an equation from those two pieces of information >_>
Hates Fun🤔
conCentrate9
Profile Blog Joined December 2007
United States438 Posts
Last Edited: 2008-10-17 00:04:56
October 17 2008 00:04 GMT
#3
Yeah. The line isn't tangent. Thanks. The point (2, -3) isn't on the parabola, the lines pass through it.
paper
Profile Blog Joined September 2004
13196 Posts
Last Edited: 2008-10-17 00:16:12
October 17 2008 00:09 GMT
#4
oh, right, i assumed it was on the parabola sorry

lemme see
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Dead9
Profile Blog Joined February 2008
United States4725 Posts
October 17 2008 00:22 GMT
#5
(y + 3) = m(x - 2)
y + 3 = (2x+ 1)(x-2)
y + 3 = 2x^2 - 3x - 2
y = 2x^2 - 3x - 5

2x^2 - 3x - 5 = x^2 + x
x^2 - 4x - 5 = 0
(x - 5)(x + 1)
I think you got it from here?
BlackStar
Profile Blog Joined July 2007
Netherlands3029 Posts
October 17 2008 00:28 GMT
#6
f(x)'= f(x) + 3 / x - 2
paper
Profile Blog Joined September 2004
13196 Posts
October 17 2008 00:31 GMT
#7
cool, dead9 wrote it up for me.

basically, you want the slope of Y at the tangential point to equal the slope of the line from (2, -3) to that point.

you know the eqn for the slope of Y [or 2x+1], and you find the slope from the point to the tangent point with (y+3)/(x-2) [which you get from the (2, -3)], so you set them equal.

y+3 / x-2 = 2x + 1, and you know that y = x^2 + x, so you substitute it. solving for X gives you 5 and -1, which are the x-coordinates of where the tangent lines hit Y.
Hates Fun🤔
micronesia
Profile Blog Joined July 2006
United States24785 Posts
Last Edited: 2008-10-17 00:52:25
October 17 2008 00:52 GMT
#8
TL seriously needs a small section devoted to hw help lol

+ Show Spoiler +
This has been proposed and dismissed several times on the grounds that tl has nothing to do with hw and should not give that topic special treatment. However, nowadays pretty much all hw help is allowed via blogs, yet I think it should just be separate from blogs to relieve some of the burden on the blogs section. A small link somewhere to a hw section along with a rule that you can't request hw help in the blogs section would do two things. One is, it would give people a place to request hw help that wasn't cumbersome, and two it would discourage people a bit from asking for help since the hw help section would no doubt receive less attention.

That or moderate hw help to some extent.
ModeratorThere are animal crackers for people and there are people crackers for animals.
AngryLlama
Profile Joined September 2005
United States1227 Posts
October 17 2008 00:55 GMT
#9
(x + 4) = ' (x) / - 2 x3 + f(x) - ' 5
conCentrate9
Profile Blog Joined December 2007
United States438 Posts
October 17 2008 06:04 GMT
#10
Thanks guys.
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