Question

determine all intervals on which the graph of f is decreasing.

Explanation:

Step1: Recall decreasing function definition

A function \( f(x) \) is decreasing on an interval if, as \( x \) increases, \( f(x) \) decreases. Visually, the graph falls from left to right on that interval.

Step2: Analyze the graph's slope

• From \( x = -1 \) to \( x = 4 \) (or the vertex around \( x = -1 \) to the minimum at \( x = 4 \)? Wait, let's check the graph: The peak is around \( x = -1 \) (since at \( x=-1 \), the y - value is 3, then it decreases until the minimum around \( x = 4 \), then increases again. Also, on the left, from \( x=-\infty \) to \( x=-6 \)? Wait no, looking at the graph: from \( x=-6 \) to \( x=-1 \), the graph is increasing (goes up from \( x=-6 \) (where it touches the x - axis) to \( x=-1 \) (peak at y = 3)). Then from \( x=-1 \) to \( x = 4 \) (the minimum point), the graph is decreasing (goes down from \( x=-1 \) (y = 3) to \( x = 4 \) (y=-2)). Then from \( x = 4 \) to \( x=\infty \), it's increasing. Wait, also, on the far left, from \( x=-\infty \) to \( x=-6 \), the graph is increasing? Wait no, the leftmost part: the graph comes from the bottom left (going up) to \( x=-6 \) (where it touches the x - axis). Then from \( x=-6 \) to \( x=-1 \), it's increasing (up to the peak at \( x=-1 \)). Then from \( x=-1 \) to \( x = 4 \), it's decreasing (down to the minimum at \( x = 4 \)). Then from \( x = 4 \) to \( x=\infty \), it's increasing.

Wait, let's re - examine the x - values: The critical points (where the slope changes) are at \( x=-1 \) (local maximum) and \( x = 4 \) (local minimum). So the function is decreasing when the x - values are between the local maximum and local minimum. So the interval where the graph is decreasing is from \( x=-1 \) to \( x = 4 \)? Wait, no, looking at the graph: when \( x \) goes from - 1 to 4, the y - value decreases (from 3 to - 2). Also, is there another interval? Wait, the left part: from \( x=-\infty \) to \( x=-6 \), is it decreasing? No, because from \( x=-\infty \) to \( x=-6 \), the graph is going up (since at \( x=-9 \), the y - value is low, and at \( x=-6 \), it's at the x - axis, so it's increasing. Then from \( x=-6 \) to \( x=-1 \), it's increasing (up to the peak at \( x=-1 \)). Then from \( x=-1 \) to \( x = 4 \), it's decreasing (down to the minimum at \( x = 4 \)). Then from \( x = 4 \) to \( x=\infty \), it's increasing.

Wait, maybe I made a mistake. Let's look at the x - axis: the graph has a local maximum at \( x=-1 \) (y = 3) and a local minimum at \( x = 4 \) (y=-2). So the function is decreasing on the interval where \( x \) is between - 1 and 4, because as \( x \) increases from - 1 to 4, \( f(x) \) decreases.

Wait, but let's check the coordinates: at \( x=-1 \), \( y = 3 \); at \( x = 0 \), \( y = 2 \) (decreasing); at \( x = 1 \), \( y = 0 \) (decreasing); at \( x = 2 \), \( y=-1 \) (decreasing); at \( x = 3 \), \( y=-2 \) (decreasing); at \( x = 4 \), \( y=-2 \) (wait, maybe the minimum is at \( x = 4 \)). So the interval where the function is decreasing is \( (-1,4) \)? Wait, no, the domain: the local maximum is at \( x=-1 \), so the interval of decrease is from \( x=-1 \) to \( x = 4 \), so in interval notation, \( (-1,4) \).

Wait, maybe I misread the graph. Let's check again: The graph crosses the y - axis at (0,2), then goes down to the x - axis at (1,0), then down to a minimum, then up. The local maximum is at \( x=-1 \) (y = 3), then it decreases until \( x = 4 \) (the minimum point), then increases. So the interval where \( f(x) \) is decreasing is \( (-1,4) \).

Answer:

The interval on which the graph of \( f \) is decreasing is \( (-1, 4) \) (or in some cases, if we consider the exact graph, maybe \( (-1, 4) \) is the interval. Wait, maybe the correct interval is \( (-1, 4) \).