How do you find the vertex of a parabola f(x) = x^2 - 2x - 3f(x)=x2−2x−3?

1 Answer
Grégory V.
Jul 10, 2015

The vertex of f(x)f(x) is -4−4 when x=1x=1 graph{x^2-2x-3 [-8, 12, -8.68, 1.32]}

Explanation:

Let a,b,ca,b,c, 3 numbers with a!=0a≠0

Let pp a parabolic function such as p(x) = a*x^2 + b*x + cp(x)=a⋅x2+b⋅x+c

A parabola always admit a minimum or a maximum (= his vertex).

We have a formula to find easily the abscissa of a vertex of a parabola :

Abscissa of vertex of p(x) = -b/(2a)p(x)=−b2a



Then, the vertex of f(x)f(x) is when (-(-2))/2=1−(−2)2=1

And f(1) = 1 - 2 - 3 = -4f(1)=1−2−3=−4


Therefore the vertex of f(x)f(x) is -4−4 when x=1x=1

Because a>0a>0 here, the vertex is a minimum.

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