How do you find the points where the graph of the function #y=x^4 - 8x^2 +2# has horizontal tangents and what is the equation?

Question

Emilia Aguilar · Accepted Answer

Horizontal tangents are at points #(0,2)#,
#(4/sqrt3,-110/9)# and #(-4/sqrt3,-110/9)#.

For #y=f(x)#, horizontal tangents will appear where #f'(x)=0#. As #y=x^4-8x^2+2#, #f'(x)=3x^3-16x=x(3x^2-16)=x(xsqrt3-4)(xsqrt3+4)# Hence #f'(x)=0# is at #x=0# and #x=4/sqrt3# and #x=-4/sqrt3# Equation of a horizontal line parallel to #x#-axis is of type #y=a#, hence let us find value of #y# at these values of #x#. At #x=0#, #y=0^4-8*0^2+2=2# at #x=4/sqrt3#, #y=(4/sqrt3)^4-8*(4/sqrt3)^2+2=256/9-128/3+2=-110/9# and at #x=-4/sqrt3#, #y=(-4/sqrt3)^4-8*(-4/sqrt3)^2+2=256/9-128/3+2=-110/9# Hence horizontal tangents are #y=2# and #y=-110/9# i.e. #9y+110=0# at points #(0,2)#, #(4/sqrt3,-110/9)# and #(-4/sqrt3,-110/9)#. graph{x^4-8x^2+2 [-5, 5, -20, 20]}

Evelyn Agard · Answer

undefined To find the points where the graph of the function y=x^4 - 8x^2 +2 has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of the function: y' = 4x^3 - 16x.

Next, we set the derivative equal to zero and solve for x: 4x^3 - 16x = 0.

Factoring out 4x, we get: 4x(x^2 - 4) = 0.

Setting each factor equal to zero, we have two possibilities: 4x = 0 or x^2 - 4 = 0.

Solving for x, we find x = 0 or x = ±2.

Therefore, the points where the graph of the function has horizontal tangents are (0, 2) and (0, -2).