Question

use the graph of f(x) to complete the table.
x | f(x)
-1 |
0 | 2

Explanation:

Step1: Analyze the graph's symmetry

The graph is a parabola symmetric about the y - axis (since it's symmetric with respect to \(x = 0\)). So, \(f(-x)=f(x)\) for all \(x\) in the domain.

Step2: Find \(f(-1)\) and \(f(1)\)

For \(x=-1\), since the graph is symmetric about the y - axis, the value of \(f(-1)\) is equal to \(f(1)\). First, let's assume the equation of the parabola. The vertex of the parabola is at \((0,2)\), and it opens downward. The general form of a parabola with vertex \((h,k)\) is \(y = a(x - h)^2+k\). Here, \(h = 0\), \(k = 2\), so \(y=ax^{2}+2\). We can find \(a\) by looking at another point. Let's take the point \((1,y)\) or \((-1,y)\). From the graph, when \(x = 1\), let's see the y - value. Alternatively, since the parabola is symmetric, we can observe that the graph passes through points that satisfy the symmetry. Let's check the value at \(x = 1\). Since the vertex is at \((0,2)\) and the parabola opens downward, let's assume the equation is \(y=-2x^{2}+2\) (we can verify: when \(x = 0\), \(y = 2\); when \(x = 1\), \(y=-2(1)^{2}+2=0\)? Wait, no, maybe my assumption of \(a\) is wrong. Wait, looking at the graph, when \(x = 1\), what is the y - value? Let's look at the grid. The vertex is at \((0,2)\). Let's move 1 unit to the right (x = 1) and see the y - coordinate. From the graph, the parabola at \(x = 1\) seems to have a y - value of, let's check the symmetry. Wait, the table has \(x=-1\), \(x = 0\) (which is 2), and \(x = 1\). Since the graph is symmetric about the y - axis, \(f(-1)=f(1)\). Let's look at the graph: when \(x = 1\), the point on the parabola: let's see the coordinates. The parabola has vertex \((0,2)\) and passes through, say, when \(x = 1\), what's the y? Wait, maybe the equation is \(y=-2x^{2}+2\). Wait, when \(x = 1\), \(y=-2(1)+2 = 0\)? No, that doesn't match. Wait, maybe the scale is different. Wait, the grid lines: each grid is 1 unit. The vertex is at \((0,2)\). Let's look at the point where \(x = 1\): the parabola at \(x = 1\) is at y - value equal to, let's see, from the graph, when \(x = 1\), the y - coordinate is 0? Wait, no, maybe I made a mistake. Wait, the table has \(x = 0\) with \(f(0)=2\). Let's look at the graph again. The parabola goes from \((0,2)\) downwards. Let's take \(x = 1\): moving 1 unit to the right from the vertex, the y - value. Let's count the grid. From \(x = 0\), y = 2. At \(x = 1\), the y - value: let's see the line of the parabola. It seems that at \(x = 1\), the y - value is 0? Wait, no, maybe the equation is \(y=-2x^{2}+2\) is incorrect. Wait, maybe the correct equation is \(y=-x^{2}+2\). Then when \(x = 1\), \(y=-1 + 2=1\)? No, that doesn't match. Wait, maybe I should just look at the graph's symmetry. Since the graph is symmetric about the y - axis, \(f(-1)=f(1)\). Let's look at the table: for \(x=-1\), we need to find \(f(-1)\), and for \(x = 1\), find \(f(1)\). From the graph, when \(x = 1\), the point on the parabola: let's see, the vertex is at \((0,2)\), and the parabola opens downward. Let's check the value at \(x = 1\): if we move 1 unit to the right from the vertex, the y - coordinate. Let's assume that the parabola has a point at \(x = 1\) with \(y = 1\)? No, wait, maybe the answer is that \(f(-1)=f(1)\), and from the symmetry, since at \(x = 0\) it's 2, at \(x = 1\) and \(x=-1\), let's see the graph. Wait, maybe the equation is \(y=-2x^{2}+2\) is wrong. Wait, let's check the point \((1,0)\): if \(x = 1\), \(y = 0\), then \(0=a(1)^{2}+2\), so \(a=-2\). So the equation is \(y=-2x^{2}+2\). Then for \(x=-1\), \(f(-1)=-2(-1)^{2}+2=-2 + 2=0\)? Wait, but when \(x = 0\), \(f(0)=2\), which matches. Then for \(x = 1\), \(f(1)=-2(1)^{2}+2=0\). So \(f(-1)=0\) and \(f(1)=0\).

Wait, let's confirm with the graph. The graph at \(x = 1\): the point \((1,0)\) is on the parabola? Let's see the grid. The vertex is at \((0,2)\). Moving 1 unit to the right (x = 1), the y - coordinate: from (0,2) down to (1,0)? Yes, that seems to fit. So for \(x=-1\), \(f(-1)=0\) (because of symmetry, same as \(x = 1\)), and for \(x = 1\), \(f(1)=0\).

Step1: Find \(f(-1)\)

Since the graph of \(f(x)\) is symmetric about the y - axis (parabola opening downward with vertex at \((0,2)\)), \(f(-1)=f(1)\). The equation of the parabola is \(y=-2x^{2}+2\) (vertex form \(y=a(x - 0)^{2}+2\), and using the point \((1,0)\) to find \(a\): \(0=a(1)^{2}+2\Rightarrow a=-2\)). Substitute \(x=-1\) into \(y=-2x^{2}+2\): \(y=-2(-1)^{2}+2=-2 + 2=0\).

Step2: Find \(f(1)\)

Using the symmetry of the parabola (symmetric about the y - axis), \(f(1)=f(-1)\). So \(f(1)=0\) (since \(f(-1)=0\)). Also, we can substitute \(x = 1\) into \(y=-2x^{2}+2\): \(y=-2(1)^{2}+2=0\).

Answer:

For \(x=-1\), \(f(-1)=\boldsymbol{0}\); for \(x = 1\), \(f(1)=\boldsymbol{0}\) (assuming the table has \(x=-1\), \(x = 0\), \(x = 1\) and we need to fill \(f(-1)\) and \(f(1)\); if the table only needs \(f(-1)\) and \(f(1)\) as per the given table structure with \(x=-1\), \(0\) (2), \(1\), then \(f(-1)=0\) and \(f(1)=0\))