How do you graph the parabola #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#

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

#y=(0-3)^2+5=9+5=14#

Its Y-intercept #(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 # to #6#

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