1. **State the problem:** We are given the function $f(x) = x^2 - 2x - 3$ and a table of $x$ values: 0, 1, -1, 3, -2. We need to find the corresponding $y$ values and understand the graph of this quadratic function. 2. **Formula and rules:** The function is a quadratic polynomial of the form $f(x) = ax^2 + bx + c$ where $a=1$, $b=-2$, and $c=-3$. The graph is a parabola opening upward because $a > 0$. 3. **Calculate $y$ values for each $x$:** - For $x=0$: $$f(0) = 0^2 - 2\cdot0 - 3 = -3$$ - For $x=1$: $$f(1) = 1^2 - 2\cdot1 - 3 = 1 - 2 - 3 = -4$$ - For $x=-1$: $$f(-1) = (-1)^2 - 2\cdot(-1) - 3 = 1 + 2 - 3 = 0$$ - For $x=3$: $$f(3) = 3^2 - 2\cdot3 - 3 = 9 - 6 - 3 = 0$$ - For $x=-2$: $$f(-2) = (-2)^2 - 2\cdot(-2) - 3 = 4 + 4 - 3 = 5$$ 4. **Summary of points:** The points are $(0,-3)$, $(1,-4)$, $(-1,0)$, $(3,0)$, and $(-2,5)$. 5. **Vertex and axis of symmetry:** The vertex of a parabola $f(x) = ax^2 + bx + c$ is at $x = -\frac{b}{2a} = -\frac{-2}{2\cdot1} = 1$. The vertex $y$ value is $f(1) = -4$, so the vertex is at $(1,-4)$. 6. **Graph behavior:** The parabola opens upward with vertex at $(1,-4)$ and crosses the $x$-axis at $x=-1$ and $x=3$ (roots). **Final answer:** The $y$ values for the given $x$ values are $-3$, $-4$, $0$, $0$, and $5$ respectively. The parabola has vertex $(1,-4)$ and roots at $x=-1$ and $x=3$.