What are the intercepts for #y=x^2+x+1#?

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Answer 1 · 2016-07-23T07:06:56

It has a #y# intercept #(0, 1)# and no #x# intercepts.

If #x=0# then #y = 0 + 0 + 1 = 1#.

So the intercept with the #y# axis is #(0, 1)#

Notice that:

#x^2+x+1 = (x+1/2)^2+3/4 >= 3/4# for all Real values of #x#

So there is no Real value of #x# for which #y=0#.

In other words there is no #x# intercept.

graph{(y-(x^2+x+1))(x^2+(y-1)^2-0.015)=0 [-5.98, 4.02, -0.68, 4.32]}