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

##### 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 $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$. 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 $2 x$, deriving $4 x$ gives 4, and deriving $- 2$ (as any other constant) gives zero. The first derivative of ${x}^{2} + 4 x - 2$ is thus $2 x + 4$.

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

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