How do you identify the important parts of #y = x^2 + x + 1# to graph it?

1 Answer
Anjali G
Dec 23, 2016

Vertex at #(-1/2,3/4)#
Y intercept at #(0,1)#
No x intercepts graph{x^2+x+1 [-10, 10, -5, 5]}

Explanation:

#y=x^2+x+1#

Vertex:
Complete the square into vertex form #y=a(x-h)+k#
#y=(x+1/2)^2+3/4#

Intercepts:
The y intercept is at #(0,1)#. We know this because if we plug in 0 in for x, the output is 1.

There are no x intercepts because the parabola is facing up and has a vertex above the x axis.

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