Application of Differentiation Prove that if the curve $y = x^3 + px + q$ is tangent to the x-axis, then 
$$4p^3 + 27q^2 = 0$$
I differentiated $y$ and obtained the value $3x^2 + p$. If the curve is tangent to the x-axis, it implies that $x=0$ (or is it $y = 0$?). How do I continue to prove the above statement? Thanks. 
If I substitute in $x=0$, I will obtain $y= q$? Are my above steps correct? Please guide me. Thank you so much!

The curve $y=f(x)$ is tangent to the $x$ axis if there exists some $x_0$ such that $$f(x_0) = 0$$ and $$f'(x_0) = 0$$
From the equation $f'(x_0) = 0$, you can see that only two possible values of $x_0$ are possible. Using them in the equation $f(x_0) = 0$ yields your result.
