1. **State the problem:** We analyze the function $f$ given by a parabola opening upwards with vertex at $(4,-1)$, x-intercepts at $x=2$ and $x=6$, and y-intercept at $y=8$. 2. **Domain:** The domain of any parabola is all real numbers, so the domain is $$(-\infty, \infty).$$ 3. **Range:** Since the parabola opens upwards and the vertex is the minimum point at $y=-1$, the range is $$[-1, \infty).$$ 4. **X-intercepts:** The parabola crosses the x-axis where $f(x)=0$. Given intercepts at $x=2$ and $x=6$, the x-intercepts are $$2 \text{ and } 6.$$ 5. **Y-intercept:** The parabola crosses the y-axis at $x=0$. Given $f(0)=8$, the y-intercept is $$8.$$ 6. **Intervals of increase and decrease:** - The vertex is at $x=4$. - For $x<4$, the parabola is decreasing. - For $x>4$, the parabola is increasing. So, - Increasing on $$(4, \infty)$$ - Decreasing on $$(-\infty, 4)$$ 7. **Intervals where $f$ is constant:** A parabola is never constant on any interval, so there are none. 8. **Relative minimum:** The vertex at $x=4$ is the relative minimum point. - The relative minimum value is $$f(4) = -1.$$ 9. **Evaluate $f(-2)$:** Since the parabola passes through $(0,8)$ and has vertex $(4,-1)$, we can find $f(-2)$ by using the quadratic formula or symmetry. The parabola has roots at $2$ and $6$, so it can be written as $$f(x) = a(x-2)(x-6).$$ Using the vertex $(4,-1)$: $$-1 = a(4-2)(4-6) = a(2)(-2) = -4a \implies a = \frac{1}{4}.$$ So, $$f(x) = \frac{1}{4}(x-2)(x-6) = \frac{1}{4}(x^2 - 8x + 12).$$ Calculate $f(-2)$: $$f(-2) = \frac{1}{4}((-2)^2 - 8(-2) + 12) = \frac{1}{4}(4 + 16 + 12) = \frac{32}{4} = 8.$$ 10. **Solve $f(x) = 3$:** $$\frac{1}{4}(x^2 - 8x + 12) = 3 \implies x^2 - 8x + 12 = 12.$$ Simplify: $$x^2 - 8x + 12 - 12 = 0 \implies x^2 - 8x = 0.$$ Factor: $$x(x - 8) = 0 \implies x = 0 \text{ or } x = 8.$$ 11. **Even, odd, or neither:** - Check $f(-x)$: $$f(-x) = \frac{1}{4}((-x)^2 - 8(-x) + 12) = \frac{1}{4}(x^2 + 8x + 12).$$ - Compare with $f(x)$: $$f(x) = \frac{1}{4}(x^2 - 8x + 12).$$ Since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$, the function is **neither even nor odd**. **Final answers:** - (a) Domain: $$(-\infty, \infty)$$ - (b) Range: $$[-1, \infty)$$ - (c) X-intercepts: $$2, 6$$ - (d) Y-intercept: $$8$$ - (e) Increasing on $$(4, \infty)$$ - (f) Decreasing on $$(-\infty, 4)$$ - (g) Constant on $$\varnothing$$ (none) - (h) Relative minimum at $$x=4$$ - (i) Relative minimum value: $$-1$$ - (j) $f(-2) = 8$ - (k) $f(x) = 3$ at $$x=0, 8$$ - (l) Function is **neither even nor odd**.