Question

determine the domain on which the following function is increasing.

Explanation:

Step1: Identify the vertex x - coordinate

The graph is a parabola opening downward (since the coefficient of \(x^{2}\) is negative, as the parabola opens down). The vertex of a parabola \(y = ax^{2}+bx + c\) is at \(x=-\frac{b}{2a}\), and for a downward - opening parabola, 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 maximum point) has an x - coordinate of \(x = 3\).

Step2: Determine the interval of increase

To find where the function is increasing, we look at the part of the graph where as \(x\) increases, \(y\) also increases. For a downward - opening parabola, the function is increasing on the interval \((-\infty, 3)\) (or in terms of the domain from the leftmost point to the x - coordinate of the vertex). But we also need to consider the domain of the function based on the graph. The graph intersects the x - axis at \(x = 0\) (wait, no, looking at the graph, the roots are at \(x = 0\) (origin) and \(x=5\)? Wait, no, let's re - examine the graph. The parabola passes through the origin \((0,0)\) and another root at \(x = 5\)? Wait, no, the x - intercepts: when \(y = 0\), the points are \(x = 0\) and \(x=5\)? Wait, no, looking at the graph, the parabola crosses the x - axis at \(x = 0\) (the origin) and at \(x = 5\)? Wait, no, the vertex is at \(x = 3\). Wait, the general form of a parabola with roots at \(x_1\) and \(x_2\) is \(y=a(x - x_1)(x - x_2)\). If the roots are \(x = 0\) and \(x = 5\), then \(y=a(x)(x - 5)=ax^{2}-5ax\). The vertex of this parabola is at \(x=\frac{0 + 5}{2}=\frac{5}{2}=2.5\)? Wait, maybe my initial observation was wrong. Wait, looking at the graph, the vertex (the peak) is at \(x = 3\) (since between \(x = 0\) and \(x = 5\), the peak is at \(x = 3\)).

Wait, the correct way: for a function \(y = f(x)\), we say that \(f(x)\) is increasing on an interval \((a,b)\) if for any two points \(x_1,x_2\in(a,b)\) with \(x_1<x_2\), we have \(f(x_1)<f(x_2)\). From the graph, we can see that as we move from left to right (increase \(x\)) towards the vertex (the maximum point), the \(y\) - value is increasing. The vertex has an x - coordinate of \(x = 3\). Also, the domain of the function (the set of all \(x\) - values for which the function is defined) from the graph: the function is a parabola, and we can see that the function is defined for all real numbers, but the increasing interval is from \(-\infty\) to \(3\)? Wait, no, looking at the graph, the left - hand side of the parabola (from the left end towards the vertex) is increasing. Wait, the parabola has a minimum? No, it's a downward - opening parabola, so it has a maximum. Wait, no, the graph: the two ends of the parabola are going down (since the arrows are pointing down), so it's a downward - opening parabola, so it has a maximum. So the function is increasing when we move from left to right towards the maximum (vertex) and decreasing when we move from the maximum to the right.

From the graph, the vertex (the maximum point) has an x - coordinate of \(x = 3\). So the function is increasing on the interval \((-\infty, 3)\)? Wait, no, the graph starts from the bottom left (going up), reaches the vertex at \(x = 3\), then goes down to the bottom right. Wait, the domain of the function (the x - values) for which the function is increasing: we can see that as \(x\) increases from \(-\infty\) to \(3\), the \(y\) - value increases. But we also need to consider the actual domain of the function based on the graph. Wait, the graph is a parabola, so the domain is all real numbers, but the increasing interval is from \(-\infty\) to \(3\)? Wait, no, looking at the graph, the left - most part: the parabola comes from the bottom left (as \(x\) increases from \(-\infty\) to \(3\), \(y\) increases), and then from \(x = 3\) to \(+\infty\), \(y\) decreases. But also, the graph intersects the x - axis at \(x = 0\) and \(x = 5\). Wait, maybe the function is defined for all real numbers, but the increasing interval is \((-\infty, 3)\)? Wait, no, let's check the slope. The slope of the tangent line to the parabola: for a downward - opening parabola \(y=ax^{2}+bx + c\) with \(a<0\), the derivative \(y^\prime=2ax + b\). The function is increasing when \(y^\prime>0\), i.e., \(2ax + b>0\). Since \(a<0\), \(x<-\frac{b}{2a}\), which is the x - coordinate of the vertex.

From the graph, the vertex is at \(x = 3\), so the function is increasing on the interval \((-\infty, 3)\). But we also need to consider the domain of the function. Wait, the graph is a parabola, so the domain is all real numbers, but the function is increasing for \(x<3\). But looking at the graph, the parabola starts from the bottom left, goes up through the origin, reaches the vertex at \(x = 3\), then goes down to the bottom right, crossing the x - axis at \(x = 5\). Wait, maybe the correct interval is from \(x = 0\) to \(x = 3\)? No, that's not right. Wait, no, the function is defined for all real numbers, but the increasing part is from \(-\infty\) to \(3\). But in the context of the graph, maybe the domain of the function (the part we are considering) is from \(x = 0\) to \(x = 5\)? No, the parabola extends to \(\pm\infty\) (since the arrows are pointing down, meaning as \(x\to\pm\infty\), \(y\to-\infty\)).

Wait, I think I made a mistake in the root identification. Let's look at the graph again: the parabola passes through \((0,0)\), \((5,0)\), and the vertex at \((3,2)\) (approximate). So the roots are \(x = 0\) and \(x = 5\), so the axis of symmetry is \(x=\frac{0 + 5}{2}=2.5\)? Wait, that contradicts the earlier observation. Maybe the graph is not a parabola with roots at \(0\) and \(5\). Wait, the y - intercept is at \((0,- 2)\) (wait, no, the graph passes through the origin \((0,0)\) and \((0,-2)\)? No, the graph crosses the y - axis at \((0,-2)\)? Wait, no, looking at the graph: the point at \(x = 0\) is \((0,0)\) (since it's on the x - axis) and the y - intercept is at \((0,-2)\)? No, the graph goes through \((0,0)\) (x - intercept) and \((0,-2)\) (y - intercept)? No, that's impossible. Wait, the graph: when \(x = 0\), \(y = 0\) (it's on the x - axis), and when \(x = 1\), \(y\) is positive, when \(x = 5\), \(y = 0\). The vertex is at \(x = 3\), \(y = 2\) (approximate). So the function is \(y=-\frac{2}{9}(x - 3)^{2}+2\) (since the vertex is \((3,2)\) and it passes through \((0,0)\): \(0=-\frac{2}{9}(0 - 3)^{2}+2=-\frac{2}{9}\times9 + 2=-2 + 2=0\), which works). Then the derivative \(y^\prime=-\frac{4}{9}(x - 3)\). The function is increasing when \(y^\prime>0\), i.e., \(-\frac{4}{9}(x - 3)>0\), which implies \(x-3<0\), so \(x<3\).

So the function is increasing on the interval \((-\infty, 3)\). But in the context of the graph, since the graph is a parabola with roots at \(x = 0\) and \(x = 5\), and the vertex at \(x = 3\), the function is increasing for all \(x\) such that \(x<3\).

Answer:

The function is increasing on the interval \((-\infty, 3)\) (or in interval notation, \(\boldsymbol{(-\infty, 3)}\)). If we consider the domain based on the visible graph (from the left - hand side up to \(x = 3\)), the domain on which the function is increasing is \(x<3\), or in interval form \((-\infty, 3)\).