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

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Answer 1 · 2018-08-10T05:36:05

Refer Explanation

Given -

$y = {(x - 2)}^{2} + 1$ $y=(x-2)^2+1$

The equation is in the form $y = (x - h) + k$ $y=(x-h)+k$ , then the vertex $(h, k)$ $(h,k)$ .

In the given equation the vertex is $(2, 1)$ $(2,1)$

Its y-intercept.

At $x = 0; y = {(0 - 2)}^{2} + 1 = 4 + 1 = 5$ $x=0; y=(0-2)^2+1=4+1=5$

y-intercept $(0, 5)$ $(0,5)$

Its y - intercept $(0, 5)$ $(0,5)$ is above the vertex $(2, 1)$ $(2,1)$ the curve opens up. Hence it has no x-intercept.

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