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eroshea 2 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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1. 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?

2. eroshea

yes.. so how will i find the slope?

3. 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.

4. terenzreignz

Start by differentiating, though.

5. eroshea

dy/dx = 6x^2 - 6x -2

6. terenzreignz

Quick to the point, are we? :) Now, when x = 1, what's dy/dx?

7. eroshea

dy/dx = -2

8. 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)

9. eroshea

negative reciprocal of it.. 1/2

10. 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?

11. eroshea

hmm.. where they would intersect? but i don't know how

12. 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

13. eroshea

y = -2

14. eroshea

then i just need to use slope intercept form , right? with P(1,-2) and a slope of 1/2?

15. terenzreignz

No... that's dy/dx I meant y. You know... your original curve? :P

16. terenzreignz

y=2x^3 - 3x^2 -2x +5

17. eroshea

y = -8.. sorry

18. eroshea

wrong again!

19. 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: \[\Large y = \color{blue}m(x-\color{red}p)+\color{green}q\]

20. eroshea

OWH.. i got -2 again

21. eroshea

its 2 right?

22. terenzreignz

What is? Your slope? No, you were right, it's -2 for the tangent.

23. eroshea

no .. the coordinates of the point (1,2)

24. terenzreignz

Ah well then, it won't be (1,-8) but (1,2) instead :) Work with that then. :D

25. terenzreignz

\[\Large y = \color{blue}m(x-\color{red}p)+\color{green}q\]

26. eroshea

y = -2x + 4 ?

27. terenzreignz

correct :) Now the normal line? the only difference is the slope :)

28. eroshea

2y= x + 3

29. terenzreignz

and you're done :P

30. eroshea

:D thanks kiddo .. but you're smarter than me

31. terenzreignz

Nah. I've just read in advance :P

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