Average Rate Change 5545C8

1. **State the problem:** We are given a graph of the function $g(x)$, which is a downward-opening parabola with vertex approximately at $(-1, 2)$ and crossing the y-axis at about $(0, 1)$. We need to determine how the average rate of change on equal length input intervals changes and justify the answer. 2. **Recall the formula for average rate of change:** The average rate of change of a function $g(x)$ over an interval $[a, b]$ is given by $$\text{Average rate of change} = \frac{g(b) - g(a)}{b - a}$$ This represents the slope of the secant line connecting points $(a, g(a))$ and $(b, g(b))$ on the graph. 3. **Analyze the shape of the parabola:** Since $g(x)$ is a downward-opening parabola, its graph is concave down. The vertex at $(-1, 2)$ is the maximum point. 4. **Behavior of average rate of change on equal intervals:** - For intervals to the left of the vertex (where $x < -1$), the function is decreasing as $x$ increases, but the slope of the tangent line is increasing (less negative) because the parabola is concave down. - For intervals to the right of the vertex (where $x > -1$), the function is decreasing and the slope of the tangent line is decreasing (more negative). 5. **Conclusion:** On equal length intervals moving from left to right: - The average rate of change becomes less negative as we approach the vertex from the left. - At the vertex, the average rate of change is zero (since the slope of the tangent is zero). - Moving right from the vertex, the average rate of change becomes more negative. This means the average rate of change decreases on intervals to the right of the vertex and increases on intervals to the left of the vertex, reflecting the concave down shape of the parabola. **Final answer:** The average rate of change on equal length input intervals changes by increasing (becoming less negative) as we approach the vertex from the left, reaching zero at the vertex, and then decreasing (becoming more negative) as we move to the right of the vertex.