anonymous
  • anonymous
find the equation of the tangent and normal line to the curve y=2x^3 - 3x^2 -2x +5, where x =1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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terenzreignz
  • terenzreignz
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?
anonymous
  • anonymous
yes.. so how will i find the slope?
terenzreignz
  • terenzreignz
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.

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terenzreignz
  • terenzreignz
Start by differentiating, though.
anonymous
  • anonymous
dy/dx = 6x^2 - 6x -2
terenzreignz
  • terenzreignz
Quick to the point, are we? :) Now, when x = 1, what's dy/dx?
anonymous
  • anonymous
dy/dx = -2
terenzreignz
  • terenzreignz
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)
anonymous
  • anonymous
negative reciprocal of it.. 1/2
terenzreignz
  • terenzreignz
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?
anonymous
  • anonymous
hmm.. where they would intersect? but i don't know how
terenzreignz
  • terenzreignz
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
anonymous
  • anonymous
y = -2
anonymous
  • anonymous
then i just need to use slope intercept form , right? with P(1,-2) and a slope of 1/2?
terenzreignz
  • terenzreignz
No... that's dy/dx I meant y. You know... your original curve? :P
terenzreignz
  • terenzreignz
y=2x^3 - 3x^2 -2x +5
anonymous
  • anonymous
y = -8.. sorry
anonymous
  • anonymous
wrong again!
terenzreignz
  • terenzreignz
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 point-slope 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: \[\Large y = \color{blue}m(x-\color{red}p)+\color{green}q\]
anonymous
  • anonymous
OWH.. i got -2 again
anonymous
  • anonymous
its 2 right?
terenzreignz
  • terenzreignz
What is? Your slope? No, you were right, it's -2 for the tangent.
anonymous
  • anonymous
no .. the coordinates of the point (1,2)
terenzreignz
  • terenzreignz
Ah well then, it won't be (1,-8) but (1,2) instead :) Work with that then. :D
terenzreignz
  • terenzreignz
\[\Large y = \color{blue}m(x-\color{red}p)+\color{green}q\]
anonymous
  • anonymous
y = -2x + 4 ?
terenzreignz
  • terenzreignz
correct :) Now the normal line? the only difference is the slope :)
anonymous
  • anonymous
2y= x + 3
terenzreignz
  • terenzreignz
and you're done :P
anonymous
  • anonymous
:D thanks kiddo .. but you're smarter than me
terenzreignz
  • terenzreignz
Nah. I've just read in advance :P

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