Final answer:
The interval over which h(t) = (t+3)² + 5 has a negative average rate of change is to the right of the vertex of the parabola, which is at t = -3. As the function value decreases when moving left from any point right of t = -3, the slope of the secant line is negative.
Step-by-step explanation:
The question is asking to determine the interval over which the function h(t) has a negative average rate of change. Given the function h(t) = (t+3)² + 5, we need to find out where the slope of the secant line between any two points on the function is negative. This function is a parabola opening upwards, which means that the average rate of change is negative when moving from right to left, from a point on the right side of the vertex to a point on the left side of the vertex.
To find the vertex, we can use the formula for the vertex of a parabola in vertex form, which for the function h(t) is at t = -3. This is because the vertex form of a parabola is given by h(t) = a(t-h)² + k, where (h, k) is the vertex of the parabola. Since our function is h(t) = (t+3)² + 5, the vertex is at t = -3. So, the average rate of change of h(t) is negative for the interval to the right of t = -3, because as we move to the left from there (decreasing t), the function value decreases (since the parabola is opening upwards) giving us a negative slope.