Question

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

Answer 1

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

Explanation:

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]}