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]}

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