1. **Problem statement:** Find the distance between the points $(2,7)$ and $(x^2, x-1)$ rounded to 3 decimals. 2. **Formula:** The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Apply the formula:** Here, $(x_1, y_1) = (2,7)$ and $(x_2, y_2) = (x^2, x-1)$. $$d = \sqrt{(x^2 - 2)^2 + ((x - 1) - 7)^2} = \sqrt{(x^2 - 2)^2 + (x - 8)^2}$$ 4. **Simplify inside the square root:** $$(x^2 - 2)^2 = x^4 - 4x^2 + 4$$ $$(x - 8)^2 = x^2 - 16x + 64$$ 5. **Sum the expressions:** $$x^4 - 4x^2 + 4 + x^2 - 16x + 64 = x^4 - 3x^2 - 16x + 68$$ 6. **Final distance formula:** $$d = \sqrt{x^4 - 3x^2 - 16x + 68}$$ 7. **Interpretation:** The distance depends on the value of $x$. For any specific $x$, substitute and compute $d$ rounded to 3 decimals. **Example:** For $x=3$, $$d = \sqrt{3^4 - 3(3^2) - 16(3) + 68} = \sqrt{81 - 27 - 48 + 68} = \sqrt{74} \approx 8.602$$ **Answer:** The distance between $(2,7)$ and $(x^2, x-1)$ is $$d = \sqrt{x^4 - 3x^2 - 16x + 68}$$ rounded to 3 decimals for any given $x$.