How do you solve and graph the inequality #x^2+x+1>0#?
Solve f(x) = x^2 + x + 1 > 0
D = d^2 = b^2 - 4ac = 1 - 4 = -3 < 0.
When D < 0 --> f(x) has the same sign as a.
Since a > 0 --> f(x) > 0 regardless of values of x. The inequality is called "always true".
The parabola opens upward. There are no x-intercepts, because the parabola is completely above the x-axis.