Question

the graph of ( y = f(x) ) is shown below. find all values of ( x ) for which ( f(x) > 0 ).

Explanation:

Step1: Identify x-intercepts

The graph intersects the x - axis at \(x=-2\) and \(x = - 6\) (by observing the points where \(y = 0\) on the graph).

Step2: Determine where \(f(x)>0\)

For a parabola (the graph is a parabola opening downwards? Wait, no, looking at the graph, the parabola opens to the left? Wait, no, the axes: the x - axis and y - axis. Wait, the graph is a parabola that intersects the x - axis (horizontal axis) at \(x=-6\) and \(x=-2\)? Wait, no, let's re - examine. The horizontal axis is the x - axis? Wait, no, in the graph, the vertical axis is y and horizontal is x? Wait, no, the labels: the vertical arrow is y, horizontal is x. Wait, the graph of the function: when we look for \(f(x)>0\), we look for the regions where the y - value (since \(y = f(x)\)) is positive. So we need to find the x - values where the graph is above the x - axis (\(y>0\)).

Looking at the graph, the function crosses the x - axis (where \(y = 0\)) at two points. Let's find the x - coordinates of the x - intercepts. From the graph, the x - intercepts are at \(x=-6\) and \(x=-2\)? Wait, no, maybe I got the axes reversed. Wait, the horizontal axis: the numbers on the horizontal axis (let's call it the x - axis) go from - 10 to 10. The vertical axis (y - axis) also from - 10 to 10. The graph of the function: when \(y>0\), the graph is above the x - axis. So we need to find the interval of x where the graph is above the x - axis.

Wait, maybe the function is a parabola opening to the left? Because the graph seems to have a vertex and opens to the left. So for a left - opening parabola, the equation is of the form \((y - k)^2=4p(x - h)\) with \(p<0\). But in terms of finding where \(f(x)>0\), we need to find the x - values where \(y>0\) (since \(y = f(x)\)). Wait, no, \(f(x)\) is the y - value. So we need to find all x such that \(y=f(x)>0\), i.e., the graph is above the x - axis (where \(y = 0\)).

Looking at the graph, the x - intercepts (where \(y = 0\)) are at \(x=-6\) and \(x=-2\)? Wait, no, let's check the coordinates. Wait, the horizontal axis: the points where the graph crosses the horizontal (x - axis) are at x=-6 and x=-2? Wait, no, maybe the horizontal axis is the y - axis? No, the labels: the vertical arrow is y, horizontal is x. Wait, I think I made a mistake. Let's re - orient: the function is plotted with x on the horizontal and y on the vertical. So \(f(x)\) is the y - value. So we need to find all x such that \(y=f(x)>0\), i.e., the graph is above the x - axis (y = 0).

From the graph, the region where \(y>0\) is between the two x - intercepts? Wait, no, if the parabola opens to the left, then for \(y>0\), the x - values are between the two y - intercepts? No, I'm getting confused. Wait, let's look at the graph again. The graph of the function: when x is between - 6 and - 2, is the graph above the x - axis? Wait, no, maybe the x - intercepts are at x=-6 and x=-2, and the graph is above the x - axis (y>0) when x is between - 6 and - 2? Wait, no, let's see: if the parabola opens to the left, the vertex is at some point, and the graph is above the x - axis (y>0) between the two y - intercepts? No, I think I have the axes reversed.

Wait, maybe the horizontal axis is the y - axis and vertical is the x - axis. Let's re - define: let the vertical axis be x and horizontal be y. Then \(f(x)\) is the x - value, and we need to find x such that \(f(x)>0\), i.e., x>0. But the graph: if the vertical axis is x and horizontal is y, then the graph is a parabola opening downwards? No, this is getting too confusing. Wait, the original problem says "Find all values of x for which \(f(x)>0\)". So \(f(x)\) is the y - coordinate, so we need to find all x where the y - coordinate (of the point \((x,y)\) on the graph) is positive.

Looking at the graph, the points where y = 0 (x - intercepts) are at x=-6 and x=-2? Wait, no, let's look at the horizontal axis (x - axis) values. The graph crosses the x - axis (y = 0) at two points. Let's find the x - coordinates of these points. From the graph, the x - intercepts are at x=-6 and x=-2? Wait, no, maybe the x - intercepts are at x=-6 and x=-2, and the graph is above the x - axis (y>0) between x=-6 and x=-2? Wait, no, if the graph is a parabola opening upwards, but in the graph, it seems to open to the left. Wait, maybe the function is \(y^2=-4a(x - h)\), a left - opening parabola. Then, for \(y>0\), we can solve for x. But the question is about x such that \(f(x)>0\), where \(f(x)\) is the y - value. So we need to find all x where \(y=f(x)>0\), i.e., the graph is above the x - axis (y>0).

Looking at the graph, the x - values where the graph is above the x - axis (y>0) are between the two x - intercepts? Wait, no, let's take a step back. Let's identify the x - intercepts (where \(y = 0\)). From the graph, the two points where the graph crosses the x - axis (y = 0) are at x=-6 and x=-2. Wait, no, maybe the x - intercepts are at x=-6 and x=-2, and the graph is above the x - axis (y>0) for x between - 6 and - 2? Wait, no, if the graph is a parabola opening to the left, then when y>0, the x - values are less than the x - coordinate of the vertex? No, this is getting too complicated. Let's look at the graph again.

Wait, the horizontal axis: the numbers on the horizontal axis (x - axis) are from - 10 to 10. The vertical axis (y - axis) from - 10 to 10. The graph of the function: when we look for \(f(x)>0\), we look for the regions where the y - value is positive. So the graph is above the x - axis (y>0) between the two x - intercepts? Wait, no, the x - intercepts are at x=-6 and x=-2? Wait, no, maybe the x - intercepts are at x=-6 and x=-2, and the graph is above the x - axis (y>0) for x in (-6, - 2)? Wait, no, let's check the coordinates. Wait, the point where the graph crosses the x - axis (y = 0) are at x=-6 and x=-2. So between x=-6 and x=-2, the graph is above the x - axis (y>0), and outside, it's below? Wait, no, if the parabola opens to the left, then for y>0, the x - values are less than the vertex's x - coordinate. But maybe I should just look at the graph: the region where \(y>0\) (f(x)>0) is between the two x - intercepts. So the x - values are between - 6 and - 2, i.e., \(-6 < x < - 2\).

Wait, let's confirm: when x is between - 6 and - 2, the graph is above the x - axis (y>0), so \(f(x)>0\) for \(-6 < x < - 2\).

Step1: Find x - intercepts

The graph of \(y = f(x)\) intersects the x - axis (where \(y = 0\)) at two points. From the graph, the x - coordinates of these intercepts are \(x=-6\) and \(x=-2\).

Step2: Determine the region where \(f(x)>0\)

To find where \(f(x)>0\), we look for the x - values where the graph is above the x - axis (since \(y = f(x)\) and we want \(y>0\)). The graph is above the x - axis between the two x - intercepts. So the values of x for which \(f(x)>0\) are the x - values between - 6 and - 2, i.e., \(-6 < x < - 2\).

Answer:

\(-6 < x < - 2\) (or in interval notation \((-6, - 2)\))