Question

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?

Answer 1

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

Explanation:

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]}