Difficult Differentiation Problem - How Do You Do This...?

AlphaRock · #1 $Old$ 11-19-2008, 06:34 PM

This is the question/additional information copied out from my teacher:

Question: Find the equation of the line through (2,1) tangent to
y=x^3 - 6x + 5

Note: The given point (2,1) is NOT on the graph.

Tip: Use slope = (y2-y1)/(x2-x1) = dy/dx

Can anybody show how to do this?

My work:
y' = 3x^2 - 6
= 3(2)^2 - 6
= 6

(y-y1) = m(x-x1)
y-1 = 6(x-2)

My work is wrong...

Mathstud28 · #2 $Old$ 11-19-2008, 06:36 PM

Quote:

Originally Posted by AlphaRock $View Post$

Note: The given point (2,1) is NOT on the graph.

That should say it all.

AlphaRock · #3 $Old$ 11-19-2008, 06:49 PM

Quote:

Originally Posted by Mathstud28 $View Post$

That should say it all.

I'm really confused on how to do this... Can somebody show me?

The question is worth 10 marks. People in my class claim they have the answer...

Theoretically! · #4 $Old$ 11-19-2008, 07:15 PM

And there is an answer!...Dont evaluate y' by x=2 because point (2,1) is outside the locus of y..Just leave y' as it is,3x^2-6!Use in (y-y1)=m(x-x1) in which case m=y',x1=2,y1=1 and y=x^3-6x+5 ,use these then solve for x!In which case one of the solutions is x=-1!With this evaluate y' and carry on the math as normal!

AlphaRock · #5 $Old$ 11-19-2008, 08:55 PM

Quote:

Originally Posted by Theoretically! $View Post$

And there is an answer!...Dont evaluate y' by x=2 because point (2,1) is outside the locus of y..Just leave y' as it is,3x^2-6!Use in (y-y1)=m(x-x1) in which case m=y',x1=2,y1=1 and y=x^3-6x+5 ,use these then solve for x!In which case one of the solutions is x=1!With this evaluate y' and carry on the math as normal!

I'm still having troubles. Can somebody explain how to do this through their work?

AlphaRock · #6 $Old$ 11-19-2008, 11:01 PM

One of the answers is y=6x-11

The next one should have a negative slope passing through (2,1) and intercepting with one of the points on the graph of y=x^3-6x+5.

Help?