[Math] Finding tangent to the curve which is parallel to the x-axis

Question

Find out the equation of tangent to the curve
$y=x^4-4x^2+6$ parallel to $x$-axis.

Answer 1

A line parallel to the $x$-axis will have slope $m = 0$. So you need to take the first derivative, and set it equal to zero to solve for the $x$ values at which the slope of the tangent to your curve is zero.

$$y=x^4-4x^2+6 \implies y' = 4x^3 - 8x = 4x(x^2 - 2) = 4x(x+\sqrt 2)(x - \sqrt 2).\quad$$ Now read off the solutions to $$4x(x+\sqrt 2)(x - \sqrt 2) = 0$$

Using each solution $x_0$, find its corresponding $y$-coordinate(s) using your original equation. For point $(x_0, y_0)$ on a line, the corresponding equation of the line is given by $$y - y_0 = m(x - x_0)$$ In your case, slope being zero, your equation(s) will be given by $y = y_0$.