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

Jul 23, 2016

It has a $y$ intercept $\left(0 , 1\right)$ and no $x$ intercepts.

#### Explanation:

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

So the intercept with the $y$ axis is $\left(0 , 1\right)$

Notice that:

${x}^{2} + x + 1 = {\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4} \ge \frac{3}{4}$ 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]}