How do I find all the critical points of f(x)=x^2+4x-2f(x)=x2+4x2?

1 Answer
Sep 26, 2015

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}ddxxn=nxn1. 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^2x2 gives 2x2x, deriving 4x4x gives 4, and deriving -22 (as any other constant) gives zero. The first derivative of x^2+4x-2x2+4x2 is thus 2x+42x+4.

Of course, 2x+4=0 \iff 2x=-4 \iff x=-22x+4=02x=4x=2.

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