What is the range of the function #y = x^2#?

1 Answer
Amory W.
Aug 30, 2014

The range is #y>=0#.

Normally, you would complete the square and check the leading coefficient, #a#, to determine the concavity for the comparison sign. However, this function is already in vertex or standard form:

#y=(x-0)^2+0#

So the vertex is #(0,0)# and the leading coefficient is positive; this means the parabola is concave up and the vertex has the minimum value. The minimum value is the bottom of the range of the function.

Semantics is important, the vertex is not the minimum value because it is a point; it merely contains the minimum value.

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