How do you find the vertex of $f(x) = x^2 + 6x - 2$ ?

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Answer 1 · 2015-08-30T20:36:03

The vertex is $V=(-3;-11)$

The coordinates of a vertex of a parabola $f(x)=ax^2+bx+c$ are given by:

$V=(-b/(2a);(-Delta)/(4a))$

But instead of calculating $Delta$ you can also calculate $q$ as $f(p)$ i.e. substitute $p$ for $x$ in $f(x)$ .

How do you find the vertex of f(x) = x^2 + 6x - 2f(x) = x^2 + 6x - 2?