Question

the graph of a function ( h ) is shown below. find ( h(2) )

Explanation:

Step1: Understand the meaning of \( h(2) \)

To find \( h(2) \), we need to determine the \( y \)-value of the function \( h \) when \( x = 2 \). This means we look at the graph of the function and find the point where \( x = 2 \), then identify the corresponding \( y \)-coordinate.

Step2: Locate \( x = 2 \) on the graph

On the given graph, we find the vertical line corresponding to \( x = 2 \). Then we check the intersection of this vertical line with the graph of the function \( h \).

Step3: Determine the \( y \)-coordinate at \( x = 2 \)

From the graph, when \( x = 2 \), the point on the graph of \( h \) has a \( y \)-coordinate of \( 0 \)? Wait, no, let's re - examine. Wait, looking at the graph, when \( x = 2 \), the graph passes through \( (2,0) \)? Wait, no, maybe I made a mistake. Wait, the graph is a parabola. Let's check the coordinates again. Wait, the vertex is at \( (0,3) \), and it crosses the \( x \)-axis at \( x=-3 \) and \( x = 3 \)? Wait, no, when \( x = 2 \), let's see the grid. Each square is 1 unit. So when \( x = 2 \), the \( y \)-value: let's count the grid. Wait, the graph at \( x = 2 \): from the origin \( (0,0) \), moving 2 units to the right (x = 2), and the graph at that point is on the x - axis? Wait, no, maybe I misread. Wait, the graph is a parabola opening downwards with vertex at \( (0,3) \). The roots are at \( x=-3 \) and \( x = 3 \)? Wait, no, when \( x = 2 \), let's check the \( y \)-value. Wait, maybe I made a mistake. Wait, let's look at the graph again. The x - axis is the horizontal axis, y - axis vertical. At \( x = 2 \), the graph intersects the x - axis? Wait, no, when \( x = 2 \), the \( y \)-coordinate: let's see, the graph at \( x = 2 \): from the vertex \( (0,3) \), the equation of the parabola can be written as \( h(x)=-ax^{2}+3 \). Since it passes through \( (3,0) \), substituting \( x = 3 \), \( y = 0 \): \( 0=-a(3)^{2}+3\Rightarrow9a = 3\Rightarrow a=\frac{1}{3} \). So \( h(x)=-\frac{1}{3}x^{2}+3 \). Then \( h(2)=-\frac{1}{3}(2)^{2}+3=-\frac{4}{3}+3=\frac{5}{3}\)? Wait, no, that contradicts the graph. Wait, maybe the graph is different. Wait, the original graph: looking at the grid, when \( x = 2 \), the point on the graph is \( (2,0) \)? Wait, no, maybe the graph is drawn such that at \( x = 2 \), \( y = 0 \)? Wait, no, let's check the user's graph again. The user's graph: the x - axis is marked with - 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6. The y - axis with - 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6. The parabola has vertex at (0,3), and crosses the x - axis at x=-3 and x = 3? Wait, no, when x = 2, the graph is above the x - axis? Wait, maybe I misread the graph. Wait, the user's graph: let's count the squares. From x = 0 (vertex at (0,3)), moving to x = 2: each unit is 1. So at x = 2, the y - value: let's see, the parabola equation is \( y=-\frac{1}{3}x^{2}+3 \). So when x = 2, \( y=-\frac{4}{3}+3=\frac{5}{3}\approx1.67 \)? But that doesn't match the grid. Wait, maybe the graph is such that at x = 2, the y - value is 0? No, that can't be. Wait, maybe the graph is symmetric about the y - axis. So when x = 2, the y - value is the same as when x=-2. Let's check x=-2: at x=-2, what's the y - value? From the graph, at x=-2, the y - value is 1? Wait, no, maybe the grid is such that each square is 1 unit. Wait, the vertex is at (0,3). At x = 1, y = 2? Wait, no, maybe the equation is \( y=-x^{2}+3 \). Then at x = 1, y = 2; x = 2, y=-4 + 3=-1? No, that's not right. Wait, I think I made a mistake in the initial analysis. Let's look at the graph again. The graph is a parabola opening downwards with vertex at (0,3). It crosses the x - axis at x=-3 and x = 3 (since when y = 0, \( 0=-x^{2}/3 + 3\Rightarrow x^{2}=9\Rightarrow x=\pm3 \)). So at x = 2, the y - value is \( y=-\frac{4}{3}+3=\frac{5}{3}\approx1.67 \), but that's not an integer. Wait, maybe the graph is drawn with integer coordinates. Wait, maybe the graph at x = 2 is at y = 0? No, that would mean the root is at x = 2, but the vertex is at (0,3) and it should be symmetric. Wait, the user's graph: looking at the grid, when x = 2, the point on the graph is (2,0)? No, that would mean the roots are at x=-3 and x = 2, which is not symmetric. So I must have misread the graph. Wait, the original problem's graph: let's see the x - axis labels: - 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6. The y - axis: - 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6. The parabola: vertex at (0,3), and passes through (- 3,0) and (3,0). So when x = 2, we can calculate the y - value using the equation of the parabola. The general form of a parabola with vertex (h,k) is \( y=a(x - h)^{2}+k \). Here, h = 0, k = 3, so \( y=ax^{2}+3 \). Since it passes through (3,0), substitute x = 3, y = 0: \( 0=a(3)^{2}+3\Rightarrow9a=-3\Rightarrow a=-\frac{1}{3} \). So the equation is \( y =-\frac{1}{3}x^{2}+3 \). Now, substitute x = 2 into the equation: \( y=-\frac{1}{3}(2)^{2}+3=-\frac{4}{3}+3=\frac{5}{3}\approx1.67 \). But this is not an integer. Wait, maybe the graph is different. Wait, maybe the grid is such that each square is 1 unit, and at x = 2, the graph is at y = 0. But that would mean the root is at x = 2, which is not symmetric. Alternatively, maybe I made a mistake in the vertex. Wait, maybe the vertex is at (0,3), and the graph passes through (2,0). But then the other root would be at (- 2,0), not (- 3,0). So there is a contradiction. Wait, perhaps the original graph has a different scale. Wait, the user's graph: looking at the x - axis, from - 6 to 6, and y - axis from - 6 to 6. The parabola: let's count the units. From x = 0 (vertex, y = 3) to x = 3, y = 0. So the distance from x = 0 to x = 3 is 3 units, and y goes from 3 to 0. So the slope from (0,3) to (3,0) is \( \frac{0 - 3}{3-0}=-1 \). So the equation of the right - hand side of the parabola (for x≥0) is \( y=-x + 3 \). Ah! That's a linear approximation? No, a parabola, but maybe it's a V - shaped graph? No, it's a parabola. Wait, if we consider the right half of the parabola (x≥0) as a line from (0,3) to (3,0), then the equation is \( y=-x + 3 \). Then when x = 2, \( y=-2 + 3=1 \). Oh! Maybe the graph is a parabola that is approximated as a line on the right and left? Wait, that makes sense. So if we take the right - hand side of the parabola (from x = 0 to x = 3) as the line \( y=-x + 3 \), then when x = 2, \( y=-2 + 3=1 \). Wait, but a parabola is not a line, but maybe in the context of the grid, it's drawn as a line for simplicity. So if we use the line equation \( y=-x + 3 \) for x≥0, then at x = 2, y = 1. But earlier, with the parabola equation, we got \( \frac{5}{3}\approx1.67 \). There is a discrepancy. Wait, maybe the graph is actually a triangle? No, it's a parabola. Wait, let's look at the grid again. Each square is 1 unit. So at x = 1, y = 2 (since from (0,3) down 1 unit, right 1 unit: (1,2)). At x = 2, y = 1 (down 1 unit, right 1 unit from (1,2)). At x = 3, y = 0 (down 1 unit, right 1 unit from (2,1)). Ah! So the right - hand side of the graph is a line with slope - 1, from (0,3) to (3,0). So it's a linear segment on the right and left, making the graph a triangle - shaped (a symmetric triangle with vertex at (0,3) and base from (- 3,0) to (3,0)). In that case, the equation for x≥0 is \( y=-x + 3 \). So when x = 2, \( y=-2 + 3=1 \).

Answer:

\( h(2)=1 \)