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 2 = -12

Thus the tangent line is the line with slope -12 and through the point on the curve with x = 2.

Penny

 
 

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