To find the x-intercept of a function, we set the numerator of the function to zero because the x-intercept occurs where the output $y = 0$. For the given function:

$f(x) = \frac{x^2 + x + 2}{x - 1}$

Step 1: Solve for when the numerator equals zero

Set the numerator $x^2 + x + 2 = 0$: $x^2 + x + 2 = 0$

Step 2: Check if it has real roots

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 1$, and $c = 2$. Substituting into the discriminant $b^2 - 4ac$: $b^2 - 4ac = 1^2 - 4(1)(2) = 1 - 8 = -7$

Since the discriminant is negative, the equation has no real roots. Therefore, the numerator does not equal zero for any real value of $x$.

Step 3: Conclusion

The function $f(x)$ has no x-intercepts because the numerator $x^2 + x + 2$ does not have real roots.

Let me know if you need further clarifications or details! 😊

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Related Questions:

1. What are the vertical and horizontal asymptotes of $\frac{x^2 + x + 2}{x - 1}$?
2. How do you find the discriminant for a quadratic equation?
3. Can x-intercepts exist for complex roots?
4. What is the behavior of $f(x)$ near $x = 1$?
5. How does the numerator $x^2 + x + 2$ behave graphically?

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Tip:

For rational functions, always examine both the numerator (for x-intercepts) and the denominator (for vertical asymptotes) to fully understand the graph.