Question

calculating a rate of change of a quadratic function over an interval which is the rate of change for the interval between –6 and –3 on the x-axis?

Explanation:

Step1: Identify points on the interval

For \( x = -6 \), from the graph, the \( y \)-value (let's call it \( y_1 \)) is 6 (since the point is at \((-6, 6)\)). For \( x = -3 \), we need to find the \( y \)-value. Looking at the graph, when \( x = -3 \), let's check the pattern. Wait, actually, let's find the coordinates. Wait, the graph: at \( x = -6 \), \( y = 6 \); at \( x = -3 \), let's see the points. Wait, maybe I made a mistake. Wait, let's re - examine. Wait, the quadratic function: let's find the two points for \( x=-6 \) and \( x = - 3\). From the graph, when \( x=-6 \), \( y = 6\) (the top left orange dot). When \( x=-3 \), let's see the \( y \)-value. Wait, maybe the graph has a point at \( x=-3 \)? Wait, no, maybe we can use the formula for rate of change, which is \( \text{Rate of change}=\frac{y_2 - y_1}{x_2 - x_1} \), where \( (x_1,y_1) \) and \( (x_2,y_2) \) are the two endpoints of the interval.

Wait, let's correct: when \( x=-6 \), the point is \((-6,6)\) (since the dot is at \( x = - 6\), \( y = 6\)). When \( x=-3 \), let's find the \( y \)-coordinate. Wait, maybe the graph: let's see the symmetry. The parabola is symmetric about the y - axis? Wait, no, the vertex is at \((-1, - 1)\)? Wait, no, the bottom point is at \((-1,-1)\)? Wait, no, looking at the graph, the vertex is at \( x=-1 \), \( y = - 1\). Wait, maybe[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]

Answer:

Step1: Identify points on the interval

For \( x = -6 \), from the graph, the \( y \)-value (let's call it \( y_1 \)) is 6 (since the point is at \((-6, 6)\)). For \( x = -3 \), we need to find the \( y \)-value. Looking at the graph, when \( x = -3 \), let's check the pattern. Wait, actually, let's find the coordinates. Wait, the graph: at \( x = -6 \), \( y = 6 \); at \( x = -3 \), let's see the points. Wait, maybe I made a mistake. Wait, let's re - examine. Wait, the quadratic function: let's find the two points for \( x=-6 \) and \( x = - 3\). From the graph, when \( x=-6 \), \( y = 6\) (the top left orange dot). When \( x=-3 \), let's see the \( y \)-value. Wait, maybe the graph has a point at \( x=-3 \)? Wait, no, maybe we can use the formula for rate of change, which is \( \text{Rate of change}=\frac{y_2 - y_1}{x_2 - x_1} \), where \( (x_1,y_1) \) and \( (x_2,y_2) \) are the two endpoints of the interval.

Wait, let's correct: when \( x=-6 \), the point is \((-6,6)\) (since the dot is at \( x = - 6\), \( y = 6\)). When \( x=-3 \), let's find the \( y \)-coordinate. Wait, maybe the graph: let's see the symmetry. The parabola is symmetric about the y - axis? Wait, no, the vertex is at \((-1, - 1)\)? Wait, no, the bottom point is at \((-1,-1)\)? Wait, no, looking at the graph, the vertex is at \( x=-1 \), \( y = - 1\). Wait, maybe[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]