How do you graph #y=(x-5)^2#?

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Answer 1 · 2015-07-17T18:36:48

This is a vertical parabola - sort of U shaped - with vertex at #(5, 0)#, axis #x = 5# and intercept with the #y# axis at #(0, 25)#.

The vertex is at the minimum value of the function.

#(x-5)^2 >= 0# since it is the square of a real number.

#(x-5)^2 = 0# when #x-5 = 0#, that is when #x=5#.

Hence the vertex is at #(5, 0)#.

The axis is vertical, passing through the vertex, so its equation is #x=5#.

The intercept with the #y# axis occurs when #x=0#, so substitute #x=0# into the equation to find #y = (0 - 5)^2 = (-5)^2 = 25#

So the intercept is at #(0, 25)#

graph{(x-5)^2 [-39.58, 40.42, -5.28, 34.72]}