1. **State the problem:** We are given the quadratic function $f(x) = -0.1x^2 + 8x + 1$ and want to analyze it. 2. **Formula and rules:** A quadratic function is generally written as $f(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. 3. **Identify coefficients:** Here, $a = -0.1$, $b = 8$, and $c = 1$. 4. **Find the vertex:** The vertex $x$-coordinate is given by $x = -\frac{b}{2a}$. 5. **Calculate vertex $x$-coordinate:** $$x = -\frac{8}{2 \times -0.1} = -\frac{8}{-0.2} = 40$$ 6. **Find vertex $y$-coordinate by substituting $x=40$ into $f(x)$:** $$f(40) = -0.1(40)^2 + 8(40) + 1 = -0.1(1600) + 320 + 1 = -160 + 320 + 1 = 161$$ 7. **Interpretation:** The vertex is at $(40, 161)$, which is the maximum point since $a < 0$ (the parabola opens downward). 8. **Find $y$-intercept:** Set $x=0$, then $f(0) = 1$. 9. **Find $x$-intercepts:** Solve $-0.1x^2 + 8x + 1 = 0$. 10. **Multiply both sides by $-10$ to clear decimals:** $$\cancel{-0.1} \times -10 x^2 + 8 \times -10 x + 1 \times -10 = 0 \times -10$$ $$x^2 - 80x - 10 = 0$$ 11. **Use quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{80 \pm \sqrt{(-80)^2 - 4(1)(-10)}}{2(1)} = \frac{80 \pm \sqrt{6400 + 40}}{2} = \frac{80 \pm \sqrt{6440}}{2}$$ 12. **Simplify square root:** $$\sqrt{6440} \approx 80.25$$ 13. **Calculate roots:** $$x_1 = \frac{80 + 80.25}{2} = \frac{160.25}{2} = 80.125$$ $$x_2 = \frac{80 - 80.25}{2} = \frac{-0.25}{2} = -0.125$$ 14. **Summary:** - Vertex at $(40, 161)$ (maximum point) - $y$-intercept at $(0,1)$ - $x$-intercepts approximately at $x = 80.125$ and $x = -0.125$ This completes the analysis of the quadratic function $f(x) = -0.1x^2 + 8x + 1$.