The curve you're describing is given by the function $f(x) = 0.05(8x - x^2)$. Let's analyze it step by step.

1. Function Analysis:

The function can be written as: $f(x) = 0.05(8x - x^2)$ This is a quadratic function, and it can be rewritten as: $f(x) = 0.05(-x^2 + 8x)$ or equivalently: $f(x) = -0.05x^2 + 0.4x$ This equation represents a downward-opening parabola.

2. Key Features of the Curve:

• Vertex: The vertex form of a parabola is given by $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex. For this standard form, the vertex can be found using the formula $h = -\frac{b}{2a}$.

  • Here, $a = -0.05$ and $b = 0.4$.
  • The x-coordinate of the vertex is $x = -\frac{0.4}{2(-0.05)} = 4$.
  • Substituting $x = 4$ into the function to find the y-coordinate: $f(4) = -0.05(4^2) + 0.4(4) = -0.05(16) + 1.6 = -0.8 + 1.6 = 0.8$
  • So, the vertex is at $(4, 0.8)$.

• Intercepts:

  • X-intercepts: The function intersects the x-axis where $f(x) = 0$. Setting $f(x) = 0$: $0 = -0.05x^2 + 0.4x$ Factor out $x$: $x(-0.05x + 0.4) = 0$ So, $x = 0$ or $-0.05x + 0.4 = 0$. Solving for $x$ in the second factor: $-0.05x = -0.4 \implies x = 8$ The x-intercepts are at $x = 0$ and $x = 8$.

  • Y-intercept: The y-intercept occurs where $x = 0$. From the equation: $f(0) = 0$ So, the y-intercept is at the origin $(0, 0)$.

3. Curve Behavior:

• Third Quadrant: The curve enters from the third quadrant, which means for some negative values of $x$, $f(x)$ is also negative. This aligns with the parabola's nature as it approaches negative infinity.
• Crossing the Origin: As noted, the curve crosses the origin $(0, 0)$.
• Direction Change: The curve reaches its maximum at the vertex $(4, 0.8)$, after which it changes direction.
• Exiting in the First Quadrant: After the vertex, the curve continues downwards, becoming steeper and passing through the first quadrant.

Conclusion:

The curve behaves as described: entering from the third quadrant, becoming less steep as it approaches the vertex, crossing the x-axis at $x = 0$, reaching the maximum at $x = 4$, and then steeply decreasing as it exits in the first quadrant.

Would you like more details on any part of this analysis, or have any questions?

---

Here are 5 related questions you might find interesting:

1. How would the graph change if the coefficient of $x^2$ was positive?
2. What are the implications of changing the constant $0.05$ to a larger or smaller value?
3. How would the curve look if the vertex was at a different point?
4. What is the significance of the x-intercepts in relation to the graph's shape?
5. How would shifting the graph vertically or horizontally affect its position in the coordinate plane?

Tip: Always check the vertex and intercepts first when analyzing the shape and direction of a quadratic function's graph.