Question

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

Answer 1

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?

Answer 2

yes.. so how will i find the slope?

Answer 3

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.

Answer 4

Start by differentiating, though.

Answer 5

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

Answer 6

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

Answer 7

dy/dx = -2

Answer 8

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)

Answer 9

negative reciprocal of it.. 1/2

Answer 10

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?

Answer 11

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

Answer 12

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

Answer 13

y = -2

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

y = -8.. sorry

Answer 18

wrong again!

Answer 19

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

Answer 20

OWH.. i got -2 again

Answer 21

its 2 right?

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

y = -2x + 4 ?

Answer 27

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

Answer 28

2y= x + 3

Answer 29

and you're done :P

Answer 30

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

Answer 31

Nah. I've just read in advance :P