Determine the coordinates on the graph of Y=x^4-8x^2 where the tangent lines to the curve are horizontal

Question

Determine the coordinates on the graph of Y=x^4-8x^2 where the tangent lines to the curve are horizontal. I need help with this problem. I know I have to take the derivative but what is left? 

Answer 1

Horizontal lines have slope M=0

 

0 = M = Y' = f'(x) = 4x^3 - 16x

 

0 = 4x^3 - 16x

 

0 = 4x(x^2 - 4)

 

0 = 4x( x + 2)(x-2)

 

x=0 , x=-2, x=2

 

f(0) = 0  ---> (0,0)

f(2) = 2^4 - 8(2)^2 = 16 - 8*4 = 16 - 32 = -16 --> (2,-16)

 

f(-2) = (-2)^4 - 8(-2)^2 = 16 - 8(4) = 16 - 32 = -16 ---> (-2,-16)

  

   

Answer 2

First find the critical values, set the first derivative equal to zero:

 

y = x⁴ - 8x²

 

y' = 4x³ - 16x

 

0 = 4x(x² - 4)

 

0 = 4x(x+2)(x-2)  the critical values are 0, 2 and -2   by applying the 2nd derivative test, we can determine if they are relative mins, maxs or inflection points

 

y" = 12x² - 16

 

 

0 is a max, and 2/-2 are mins