Question

determine the domain on which the following function is increasing.

Explanation:

Step1: Identify the vertex's x - coordinate

The graph is a parabola opening downwards (since the ends go down). For a parabola \(y = ax^{2}+bx + c\) (opening downwards, \(a<0\)), the function is increasing to the left of the vertex and decreasing to the right of the vertex. From the graph, we can see that the vertex (the peak of the parabola) has an x - coordinate. Looking at the grid, the vertex seems to be at \(x = 5\) (assuming the grid lines are spaced by 1 unit, and the peak is at \(x = 5\) approximately).

Step2: Determine the increasing interval

A function is increasing when, as \(x\) increases, \(y\) also increases. For a downward - opening parabola, the function is increasing on the interval \((-\infty, h)\), where \(h\) is the x - coordinate of the vertex. But we also need to consider the domain of the function from the graph. The graph starts (the left - most point) and goes up to the vertex. From the graph, the function starts (the left end) and the increasing part is from the leftmost x - value (where the graph starts) up to the x - coordinate of the vertex. But looking at the x - intercepts, one x - intercept is at \(x = 3\) (approx) and the other at \(x = 7\) (approx), and the vertex is at \(x = 5\). Wait, actually, for a parabola, the axis of symmetry is \(x=h\), and the function is increasing when \(x < h\) (for downward opening). But from the graph, the function is defined (the graph exists) from, let's see the left end: the left end of the parabola (the minimum x - value where the graph is present) seems to be around \(x = 1\) (but maybe the key is the vertex). Wait, actually, the correct way is: for a quadratic function \(y=a(x - h)^{2}+k\) with \(a<0\), the function is increasing on \((-\infty, h)\). From the graph, the vertex is at \(x = 5\) (since the peak is at \(x = 5\)). But also, we need to check the domain of the function's graph. The graph starts (the left side) and goes up to \(x = 5\), then down. But looking at the x - axis, the function crosses the x - axis at \(x = 3\) and \(x = 7\), and the vertex is at \(x = 5\). Wait, maybe the graph is a parabola with roots at \(x = 3\) and \(x = 7\), so the axis of symmetry is \(x=\frac{3 + 7}{2}=5\). So the function is increasing on the interval \((-\infty, 5)\)? But no, the graph starts (the left end) at some \(x\) - value. Wait, the left end of the parabola (the part where it comes from below) starts at \(x = 1\) (maybe), but actually, for the domain where the function is increasing, we look at the x - values where as we move from left to right, the \(y\) - values increase. So from the leftmost point of the graph (where it starts) up to the vertex's x - coordinate. But since the parabola has roots at \(x = 3\) and \(x = 7\), and vertex at \(x = 5\), the function is increasing on the interval \((-\infty, 5)\)? Wait, no, the domain of the function's graph: the graph is a parabola opening downwards, with vertex at \((5, y_{peak})\), and it intersects the x - axis at \(x = 3\) and \(x = 7\). But the left end of the parabola (the part below the x - axis) starts at \(x = 1\) (maybe), but when we talk about the function increasing, we consider the x - values where the function is defined and the slope is positive. The derivative of a quadratic function \(y=-ax^{2}+bx + c\) (a>0) is \(y'=-2ax + b\). Setting \(y' > 0\), we get \(-2ax + b>0\), or \(x<\frac{b}{2a}\), which is the x - coordinate of the vertex. So in this case, the vertex is at \(x = 5\), so the function is increasing when \(x < 5\). But we also need to consider the domain of the function from the graph. The graph is present from, let's see, the left end (where the parabola starts) to the right end. But the key is that the function is increasing on the interval \((-\infty, 5)\)? Wait, no, maybe the graph is defined for all real numbers, but from the visual, the function is increasing from \(x = 1\) (the left end) to \(x = 5\)? Wait, no, the x - intercept is at \(x = 3\), so before \(x = 3\), the function is below the x - axis, but still increasing. Wait, let's re - examine: the parabola opens downward, so it has a maximum at \(x = 5\). So as \(x\) increases from \(-\infty\) to \(5\), the function is increasing, and as \(x\) increases from \(5\) to \(+\infty\), the function is decreasing. But from the graph, the left end of the parabola (the part that comes from the bottom) starts at \(x = 1\) (maybe), but mathematically, for a quadratic function, the domain is all real numbers, and the increasing interval is \((-\infty, h)\) where \(h\) is the x - coordinate of the vertex. Since the vertex is at \(x = 5\), the function is increasing on the interval \((-\infty, 5)\). But maybe the graph is such that the function is defined from \(x = 1\) to \(x = 7\) (since it intersects the x - axis at \(x = 3\) and \(x = 7\) and has a vertex at \(x = 5\)), but no, the parabola extends beyond? Wait, the arrows at the end of the parabola (the left and right ends) are going down, which means the parabola extends to \(-\infty\) on the left and \(+\infty\) on the right? No, that can't be, because a parabola opening downward has a maximum and extends to \(-\infty\) as \(x\to\pm\infty\). Wait, no, a quadratic function \(y = ax^{2}+bx + c\) with \(a<0\) has a maximum at \(x=-\frac{b}{2a}\) and as \(x\to\pm\infty\), \(y\to-\infty\). So the function is increasing on \((-\infty,-\frac{b}{2a})\) and decreasing on \((-\frac{b}{2a},+\infty)\). From the graph, the vertex (the maximum point) is at \(x = 5\), so the function is increasing on the interval \((-\infty, 5)\). But maybe the graph is drawn with the vertex at \(x = 5\), and the x - intercepts at \(x = 3\) and \(x = 7\), so the axis of symmetry is \(x = 5\), so the function is increasing on \((-\infty, 5)\).

Answer:

The function is increasing on the interval \((-\infty, 5)\) (or in the context of the graph, if we consider the visible domain, from the leftmost x - value up to \(x = 5\), but mathematically, for the quadratic function, it's \((-\infty, 5)\)).