How do you graph the parabola #2x^2 – 4x + 1# using vertex, intercepts and additional points?

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Answer 1 · 2017-05-10T20:38:59

#y= 2x^2 - 4x + 1#

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

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

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

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

This is of the form #y = a(x- p)^2 + q#. This means that the vertex is at #(p, q) = (1, -1)#.

The parabola opens upwards. Now let's consider intercepts.

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

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

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

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

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

As for the y-intercepts:

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

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

#y= 1#

You could make a table of values, but I'll leave that up to you. Here is the graph:

graph{y = 2x^2 - 4x+ 1 [-10, 10, -5, 5]}

Hopefully this helps!