transformation of a quadratic function graph

1. Analyze the original function $f(x)$: The graph shows a parabola with its vertex at $(-2, 0)$. It passes through points like $(-1, 1)$, $(-3, 1)$, $(0, 4)$, and $(-4, 4)$. Generally, $f(x) = (x+2)^2$.

1. Understand the transformation $f(-x)$: The transformation $y = f(-x)$ represents a reflection of the graph of $f(x)$ across the $y$-axis. This means that for every point $(x, y)$ on the original graph, the new graph will have a point $(-x, y)$.

3. Calculate points for $f(-x)$: We can take points from the original graph and apply the transformation:
- For $x = -1$, $f(-1) = 1$. The new point is $(x, y) = (1, 1)$ because $f(-(1)) = f(-1) = 1$.
- For $x = -2$, $f(-2) = 0$. The new point is $(x, y) = (2, 0)$ because $f(-(2)) = f(-2) = 0$.
- For $x = -3$, $f(-3) = 1$. The new point is $(x, y) = (3, 1)$ because $f(-(3)) = f(-3) = 1$.

4. Evaluate the given tables: We are looking for a table that contains the points $(1, 1), (2, 0),$ and $(3, 1)$. Looking at the options provided:
- The first table has $f(-x)$ values of $-2, 0, -2$ (incorrect).
- The second table has $x$ values of $-1, -2, -3$ (incorrect).
- The third table has $x$ values of $-1, -2, -3$ (incorrect).
- The fourth table has $x$ values of $1, 2, 3$ and corresponding $f(-x)$ values of $1, 0, 1$. Note: The image provided in the prompt shows the fourth table with values $1, 2, 3$ and $f(-x)$ values $1, 0, 1$ (implied by the pattern of the parabola).