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

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terenzreignzBest ResponseYou've already chosen the best response.2
To 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?
 9 months ago

erosheaBest ResponseYou've already chosen the best response.1
yes.. so how will i find the slope?
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.2
Magic. 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.
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.2
Start by differentiating, though.
 9 months ago

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

terenzreignzBest ResponseYou've already chosen the best response.2
And 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)
 9 months ago

erosheaBest ResponseYou've already chosen the best response.1
negative reciprocal of it.. 1/2
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.2
And 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?
 9 months ago

erosheaBest ResponseYou've already chosen the best response.1
hmm.. where they would intersect? but i don't know how
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.2
Well, 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
 9 months ago

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

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

terenzreignzBest ResponseYou've already chosen the best response.2
y=2x^3  3x^2 2x +5
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.2
LOL 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: \[\Large y = \color{blue}m(x\color{red}p)+\color{green}q\]
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.2
What is? Your slope? No, you were right, it's 2 for the tangent.
 9 months ago

erosheaBest ResponseYou've already chosen the best response.1
no .. the coordinates of the point (1,2)
 9 months ago

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

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

terenzreignzBest ResponseYou've already chosen the best response.2
correct :) Now the normal line? the only difference is the slope :)
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.2
and you're done :P
 9 months ago

erosheaBest ResponseYou've already chosen the best response.1
:D thanks kiddo .. but you're smarter than me
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.2
Nah. I've just read in advance :P
 9 months ago
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