How do you graph the parabola $y = {(x - 3)}^{2} + 5$ $y = (x - 3)^2 + 5$ using vertex, intercepts and additional points?

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Answer 1 · 2018-06-24T05:25:26

Refer explanation section

Given -

$y = {(x - 3)}^{2} + 5$ $y=(x-3)^2+5$

The equation of the parabola is already in vertex form
So, its vertex is $(3, 5)$ $(3,5)$
To find its y-intercept, put $x = 0$ $x=0$

$y = {(0 - 3)}^{2} + 5 = 9 + 5 = 14$ $y=(0-3)^2+5=9+5=14$

Its Y-intercept $(0, 14)$ $(0,14)$

Its vertex is in the first quadrant. Its y-intercept is well above the vertex. Hence the curve cannot have x-intercept.

Take a series of x values ; these values must include vertex and y-intercept. The range of values $- 1$ $-1$ to $6$ $6$

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How do you graph the parabola y = (x - 3)^2 + 5y=(x−3)2+5y = (x - 3)^2 + 5 using vertex, intercepts and additional points?