duwan
  • duwan
  • Herf-A-Holic Topic Starter
Wednesday, September 2, 2009 4:33:00 PM
can anyone help with this?

Find equation of the line tangent to y=6x^3 at the point (1,6)

a) Find slope

b) Use the point slope form to find tangent line equation.

andytv
Wednesday, September 2, 2009 4:36:00 PM
look in the back of the book you ****.
andytv
Wednesday, September 2, 2009 4:59:00 PM
I'll bite........

slope of the tangential line is f'(x) where (f(x)=6x^3).........so if I recall correctly, the slope is 18 (or 18/1).

So the line becomes y=18x-12

I could be all wrong........it's been a LONG time and I was stoned a lot in college.

duwan
  • duwan
  • Herf-A-Holic Topic Starter
Wednesday, September 2, 2009 5:05:00 PM
that's what had figured, just wanting another opinion.
HockeyDad
Wednesday, September 2, 2009 5:08:00 PM
The ciphering club is now called to order.
rfenst
Wednesday, September 2, 2009 5:08:00 PM
Me too! I went threw 8 credits of science calculus and a class in differential equations, before finally realizing I wasn't going to be a doctor. So, I finally double-majored in Finance and Business Economics.
rfenst
Wednesday, September 2, 2009 5:10:00 PM
6 for 3.
big chief
Wednesday, September 2, 2009 5:18:00 PM
Chief is considering becoming a witch-doctor--will this kind of math be required?
rfenst
Wednesday, September 2, 2009 5:20:00 PM
Will you prescribe pot?
robertknyc
Wednesday, September 2, 2009 5:41:00 PM
In the equation, "y=mx+b" m is the slope and b is the y intercept if I remember correctly.
MACS
Wednesday, September 2, 2009 5:43:00 PM
The hardest math class I ever had was algebra.

Never took geometry, trig or calculus. Of course, I'm not an engineer, so I guess that's okay.

JTokash
Wednesday, September 2, 2009 7:02:00 PM
Yep, Andy's right.

-JT

jimmy@ctconnect.com
Wednesday, September 2, 2009 7:04:00 PM
y = x^3-6x^2+9x-10
jimmy@ctconnect.com
Wednesday, September 2, 2009 7:09:00 PM
y1 = (1)^3-6(1)^2+9(1)-10 = 1-6+9-10 = -6;

turning point 1 (1, -6)

y1'' = 6(1)-12 = 6-12 = -6 > 0 max turning point

turning point 2 (3, -10)

y''2 = 6(3)-12 = 27-12 = 15 > 0 min turning point

JTokash
Wednesday, September 2, 2009 7:10:00 PM
y = x^3-6x^2+9x-10

y' = 3x^2-12x+9

y'' = 6x-12

y''' = 6

y'''' = 0

Shall I integrate it too?

-JT

JTokash
Wednesday, September 2, 2009 7:11:00 PM
Jimmy, WTF are you talking about? LOL

-JT

gringococolo
Wednesday, September 2, 2009 7:11:00 PM
^ get a room you two. Stop sending encoded love notes to each other.
JTokash
Wednesday, September 2, 2009 7:13:00 PM
.. .-.. --- ...- . -.-- --- ..- - --- ---
jimmy@ctconnect.com
Wednesday, September 2, 2009 7:14:00 PM
^ ... feel free ... knock yourself out ... but it's a much longer process and well, I really don't feel like drawing it out any further than it already has been ...

... if he can't figure that part out ... he might want to consider withdrawing from the class and pick another one ...

... and besides ... this is one of those 12 step programs that nobody really wants to have to go through ...

- it's almost (virtually) useless in real life ... all you need to know is "how" it works and understand "why" it works the way it does ... after that ... if you're going to sit around and put that much time into it ... you have no life ...

- so fee freel ... but don't be surprised if I get bored and correct you if you screw it up ... LOL

jimmy@ctconnect.com
Wednesday, September 2, 2009 7:17:00 PM
JT,

... apparently you didn't realize it, however, this is an x-cubed graph and will have more than one turning point...