Quandaries
and Queries
hi, my name is Louise. i am a senior in high school taking Ap calculus and i have a problem i need help answering. it would be greatly appreciated
the equation of the tangent line to the curve y = x3 - 6x2 at its point of inflection is
A. y = -12x + 8
B. y = -12x + 40
C. y + 12x - 8
D. y = -12x + 12
E. y = 12x - 40
Hi Louise,
If there is a point of inflection on this curve then it must be at a point where the second derivative is zero. Hence I found the first and second derivatives
y' = 3 x2 - 12 x
y'' = 6 x - 12
Thus the second derivative is zero if x = 2.
At x = 2 the value of the first derivative is
3
22 - 12
Thus the tangent line is the line with slope -12 and through the point on the curve with x = 2.
Penny