Question

How do I find all the critical points of `f(x)=x^2+4x-2`?

Answer 1

You find the critical points of a function by setting its first derivative to zero. In this case, you only need to remember the rule for deriving a power, which is `d/{dx} x^n = nx^{n-1}`. Also, you have to remember that the derivative of a sum is the sum of the derivatives, which means that we can derive this polynomial term by term.

So, deriving `x^2` gives `2x`, deriving `4x` gives 4, and deriving `-2` (as any other constant) gives zero. The first derivative of `x^2+4x-2` is thus `2x+4`.

Of course, `2x+4=0 \iff 2x=-4 \iff x=-2`.

The (only) critical point of your parabola is `(-2, -6)`, which is its vertex.