1. **State the problem:** We are given a parabola centered at the origin with vertex at $(0,5)$, opening downward, and x-intercepts approximately at $(-2.6,0)$ and $(2.6,0)$. We need to find: a. The domain of the function. b. The range of the function. c. The x-intercepts. d. The y-intercept. e. The missing function values $f(-2)$ and $f(2)$. 2. **Formula and properties:** The parabola is symmetric about the y-axis and opens downward, so its equation can be written as: $$f(x) = a x^2 + k$$ where $a < 0$ (since it opens downward) and $k$ is the vertex's y-coordinate. Given vertex $(0,5)$, the equation is: $$f(x) = a x^2 + 5$$ 3. **Find $a$ using x-intercepts:** The parabola crosses the x-axis at approximately $x = \pm 2.6$, so: $$0 = a (2.6)^2 + 5$$ $$a (2.6)^2 = -5$$ $$a = \frac{-5}{(2.6)^2} = \frac{-5}{6.76} \approx -0.74$$ 4. **Write the function:** $$f(x) = -0.74 x^2 + 5$$ 5. **Domain:** The domain of any parabola is all real numbers: $$\text{Domain} = (-\infty, \infty)$$ 6. **Range:** Since the parabola opens downward with vertex at $5$, the maximum value is $5$ and it goes down to negative infinity: $$\text{Range} = (-\infty, 5]$$ 7. **x-intercepts:** Given as approximately $(-2.6, 0)$ and $(2.6, 0)$. 8. **y-intercept:** At $x=0$, $$f(0) = -0.74 \times 0^2 + 5 = 5$$ So the y-intercept is $(0,5)$. 9. **Find missing values:** $$f(-2) = -0.74 \times (-2)^2 + 5 = -0.74 \times 4 + 5 = -2.96 + 5 = 2.04$$ $$f(2) = -0.74 \times 2^2 + 5 = -0.74 \times 4 + 5 = 2.04$$ **Final answers:** - Domain: $(-\infty, \infty)$ - Range: $(-\infty, 5]$ - x-intercepts: $(-2.6, 0)$ and $(2.6, 0)$ - y-intercept: $(0,5)$ - $f(-2) = 2.04$ - $f(2) = 2.04$