How do you graph #y = (x - 1)^2 + 2#?

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Answer 1 · 2018-05-20T17:34:24

#(1,2)# is the vertex

#y = (x - 1)^2 + 2#

This function is in vertex form:

#y = a(x -h) + k#

#(-h, k) = (1,2)# is the vertex

to find the y-intcept set x=0 and solve:

#y = (0 - 1)^2 + 2#

#y =3#

to find the x-intcepts set y=0 and solve:

#0 = (x - 1)^2 + 2#

#(x - 1)^2 = -2#

#sqrt((x - 1)^2) = +-sqrt(-2)#

#x - 1 = +-sqrt(-2)#

#x = 1+-sqrt(2)i#. so there are no real roots hence no x-intercepts.

graph{(x - 1)^2 + 2 [-17.48, 23.07, -0.65, 19.63]}