1. The problem asks to find the value of $x$ for which $f(x) = x^2 + x + 1$ equals the given points in the set $\{(-2,8), (-1,0), (1,1), (m,-1)\}$. 2. We use the function $f(x) = x^2 + x + 1$ and substitute the $x$ values from the points to check if the $y$ values match. 3. For $x = -2$: $$f(-2) = (-2)^2 + (-2) + 1 = 4 - 2 + 1 = 3$$ The point is $(-2,8)$ but $f(-2) = 3 \neq 8$, so this point is not on the function. 4. For $x = -1$: $$f(-1) = (-1)^2 + (-1) + 1 = 1 - 1 + 1 = 1$$ The point is $(-1,0)$ but $f(-1) = 1 \neq 0$, so this point is not on the function. 5. For $x = 1$: $$f(1) = 1^2 + 1 + 1 = 1 + 1 + 1 = 3$$ The point is $(1,1)$ but $f(1) = 3 \neq 1$, so this point is not on the function. 6. For $x = m$: We want $f(m) = -1$, so: $$m^2 + m + 1 = -1$$ $$m^2 + m + 1 + 1 = 0$$ $$m^2 + m + 2 = 0$$ 7. Solve the quadratic equation: $$m = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times 2}}{2 \times 1} = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm \sqrt{-7}}{2}$$ 8. Since the discriminant is negative, there is no real solution for $m$. 9. Therefore, none of the points except possibly the first question's options match the function values. 10. The question asks to select the correct option from 1) 3, 2) 4, 3) 1, 4) 2. 11. From the above, only $f(-2) = 3$ matches the value 3, so the answer is option 1) 3. Final answer: 1) 3