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eroshea
 one year ago
find the equation of the tangent and normal line to the curve y=2x^3  3x^2 2x +5, where x =1
eroshea
 one year 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
 one year 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?

eroshea
 one year ago
Best ResponseYou've already chosen the best response.1yes.. so how will i find the slope?

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

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

terenzreignz
 one year 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)

eroshea
 one year ago
Best ResponseYou've already chosen the best response.1negative reciprocal of it.. 1/2

terenzreignz
 one year 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?

eroshea
 one year ago
Best ResponseYou've already chosen the best response.1hmm.. where they would intersect? but i don't know how

terenzreignz
 one year 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

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

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

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

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

eroshea
 one year ago
Best ResponseYou've already chosen the best response.1no .. the coordinates of the point (1,2)

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

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

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

eroshea
 one year ago
Best ResponseYou've already chosen the best response.1:D thanks kiddo .. but you're smarter than me

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