Home > Algebra > Quadratic Functions and Their Graphs

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

Answer 1

Julia Adams

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]}

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Jace Anderson

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).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

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.