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?
y' = 3x^2 - 6
= 3(2)^2 - 6
(y-y1) = m(x-x1)
y-1 = 6(x-2)
My work is wrong...
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!