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anonymous
 3 years ago
find the equation of the tangent and normal line to the curve y=2x^3  3x^2 2x +5, where x =1
anonymous
 3 years ago
find the equation of the tangent and normal line to the curve y=2x^3  3x^2 2x +5, where x =1

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terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2To get a line, you need a slope, and a point, right? Or, knowing the slope of a line and a point where it passes through, that's sufficient to find the equation of the line, yes?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes.. so how will i find the slope?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Magic. Or, you can differentiate this function first... go ahead now :) Remember, the slope of the tangent line at the point would be its derivative, while the slope of the normal line would be the negative reciprocal of the derivative.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Start by differentiating, though.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Quick to the point, are we? :) Now, when x = 1, what's dy/dx?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2And that is the slope of your tangent line :) What's the slope of your normal line? (Remember, the normal line is *perpendicular* to the tangent line)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0negative reciprocal of it.. 1/2

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2And that is the slope of the normal line. So you have the slopes for both lines, you now only need a point. Any idea how to get one?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm.. where they would intersect? but i don't know how

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Well, both the tangent and the normal would intersect your curve at the point WHERE x = 1. So at the curve itself, when x = 1, what is y? :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then i just need to use slope intercept form , right? with P(1,2) and a slope of 1/2?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2No... that's dy/dx I meant y. You know... your original curve? :P

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2y=2x^3  3x^2 2x +5

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2LOL don't worry about that :) So you have a point (1 , 8) and you have a slope (2 for your tangent, 1/2 for your normal) You can't use a slope intercept form, you need the more general pointslope form. When you have a point \(\large (\color{red}p,\color{green}q)\) and a slope \(\large \color{blue}m\) then the equation of the line is: \[\Large y = \color{blue}m(x\color{red}p)+\color{green}q\]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2What is? Your slope? No, you were right, it's 2 for the tangent.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no .. the coordinates of the point (1,2)

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Ah well then, it won't be (1,8) but (1,2) instead :) Work with that then. :D

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2\[\Large y = \color{blue}m(x\color{red}p)+\color{green}q\]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2correct :) Now the normal line? the only difference is the slope :)

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2and you're done :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0:D thanks kiddo .. but you're smarter than me

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Nah. I've just read in advance :P
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