1. The problem asks to identify which quadratic function matches the given graph. 2. The options are: - $f(x) = -x^2$ - $f(x) = -x^2 + 1$ - $f(x) = 2x^2$ - $f(x) = 2x^2 + 1$ 3. Important properties of quadratic functions: - The graph of $f(x) = ax^2 + bx + c$ is a parabola. - If $a > 0$, the parabola opens upwards. - If $a < 0$, the parabola opens downwards. - The vertex form is $f(x) = a(x-h)^2 + k$ where $(h,k)$ is the vertex. 4. From the graph description: - The parabola opens upwards, so $a > 0$. - The vertex is at $(0,2)$, so the function can be written as $f(x) = a(x-0)^2 + 2 = ax^2 + 2$. 5. Check which option fits the vertex: - $f(x) = 2x^2 + 1$ has vertex at $(0,1)$, so no. - $f(x) = 2x^2$ has vertex at $(0,0)$, so no. - $f(x) = -x^2$ opens downwards, so no. - $f(x) = -x^2 + 1$ opens downwards, so no. 6. None of the options exactly match vertex at $(0,2)$, but the graph passes through $(1,4)$ and $(2,10)$. 7. Test $f(x) = 2x^2 + 1$ at $x=1$: $2(1)^2 + 1 = 3$, but graph shows 4. 8. Test $f(x) = 2x^2$ at $x=1$: $2(1)^2 = 2$, graph shows 4. 9. Test $f(x) = -x^2 + 1$ at $x=1$: $-1 + 1 = 0$, graph shows 4. 10. Test $f(x) = -x^2$ at $x=1$: $-1$, graph shows 4. 11. None of the options fit the points given exactly, but the vertex at $(0,2)$ and points $(1,4)$ and $(-1,4)$ suggest the function is $f(x) = 2x^2 + 2$. 12. Since this is not an option, the closest is $f(x) = 2x^2 + 1$ which is option 4. 13. Therefore, the best match is $\boxed{f(x) = 2x^2 + 1}$. Final answer: $f(x) = 2x^2 + 1$