How do you graph #y<(x-2)^2+6#?

1 Answer
Christian Z.
Oct 6, 2016

This is vertex form (h,k), with a dotted line and shading below curve

Explanation:

#y= (x-h)^2 + k# would give you the vertex. There are no roots as the vertex is up at (2,6).
Shade below the curve since it is less than.
Dotted line because there's no equal sign. graph{y<(x-2)^2+6 [-20, 20, -3.12, 16.88]}

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