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

It has a

Notice that:

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

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To find the intercepts of the quadratic equation y = x^2 + x + 1, set y to zero to find the x-intercepts, and set x to zero to find the y-intercept.

To find the x-intercepts, solve the equation 0 = x^2 + x + 1 for x. To find the y-intercept, substitute x = 0 into the equation y = x^2 + x + 1.

After solving, the x-intercepts are imaginary, and the y-intercept is at (0, 1). Therefore, the equation y = x^2 + x + 1 does not intersect the x-axis but intersects the y-axis at the point (0, 1).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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