1. **Stating the problem:** We are given the quadratic function $$f(x) = x^2 - 8x + 16$$ and asked to identify its graph. 2. **Formula and important rules:** A quadratic function is generally written as $$f(x) = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. - The graph of a quadratic function is a parabola. - If $a > 0$, the parabola opens upwards; if $a < 0$, it opens downwards. - The vertex of the parabola is at $$x = -\frac{b}{2a}$$. - The y-coordinate of the vertex is $$f\left(-\frac{b}{2a}\right)$$. - The parabola can be factored or completed the square to find roots or vertex form. 3. **Find the vertex:** - Here, $a = 1$, $b = -8$, $c = 16$. - Vertex $x$-coordinate: $$x = -\frac{-8}{2 \times 1} = \frac{8}{2} = 4$$. - Vertex $y$-coordinate: $$f(4) = 4^2 - 8 \times 4 + 16 = 16 - 32 + 16 = 0$$. - So, vertex is at $(4, 0)$. 4. **Rewrite the function by completing the square:** $$f(x) = x^2 - 8x + 16 = (x^2 - 8x + 16) = (x - 4)^2$$. 5. **Interpretation:** - The function is a perfect square, so the parabola touches the x-axis at $x=4$ (double root). - The parabola opens upwards (since $a=1 > 0$). - The vertex is at $(4,0)$, which is also the minimum point. 6. **Graph features:** - The parabola has its minimum at $(4,0)$. - It touches the x-axis at $x=4$. - The y-intercept is at $f(0) = 0^2 - 8 \times 0 + 16 = 16$. 7. **Conclusion:** - The graph is a parabola opening upwards with vertex at $(4,0)$ and y-intercept at $(0,16)$. **Final answer:** The graph corresponds to the function $$f(x) = (x - 4)^2$$, a parabola with vertex at $(4,0)$ opening upwards.