Final answer:
To find the points on the curve where the tangent line is horizontal, we need to find the critical points of the function y = x^4 - 2x^2 - 4. The critical points are x = 0, x = 1, and x = -1, which correspond to the points (0, -4), (1, -5), and (-1, -1) on the curve.
Step-by-step explanation:
To find the points on the curve where the tangent line is horizontal, we need to find the values of x that make the derivative of y with respect to x equal to zero. In other words, we need to find the critical points of the function y = x^4 - 2x^2 - 4.
To do this, we take the derivative of y with respect to x, which is dy/dx = 4x^3 - 4x. Setting this equal to zero and solving for x, we get 4x^3 - 4x = 0. Factoring out 4x, we have 4x(x^2 - 1) = 0. So, x = 0, x = 1, and x = -1 are the critical points.
Plugging these values into the original equation, we can find the corresponding y-values. When x = 0, y = -4. When x = 1, y = -5. And when x = -1, y = -1. Therefore, the points on the curve where the tangent line is horizontal are (0, -4), (1, -5), and (-1, -1).