1. **State the problem:** We are given the function $f(x) = x^2 - 2x + 5$ and asked to analyze it on the interval $x = 0$ to $x = 3$. 2. **Formula and rules:** This is a quadratic function of the form $ax^2 + bx + c$ where $a=1$, $b=-2$, and $c=5$. Quadratic functions graph as parabolas. Since $a > 0$, the parabola opens upward. 3. **Find the vertex:** The vertex $x$-coordinate is given by $x = -\frac{b}{2a} = -\frac{-2}{2 \times 1} = 1$. 4. **Calculate the vertex $y$-value:** Substitute $x=1$ into $f(x)$: $$f(1) = 1^2 - 2 \times 1 + 5 = 1 - 2 + 5 = 4$$ 5. **Evaluate the function at the interval endpoints:** - At $x=0$: $f(0) = 0^2 - 2 \times 0 + 5 = 5$ - At $x=3$: $f(3) = 3^2 - 2 \times 3 + 5 = 9 - 6 + 5 = 8$ 6. **Summary:** The parabola opens upward with vertex at $(1,4)$. On the interval $[0,3]$, the function values range from $5$ at $x=0$, down to $4$ at $x=1$, then up to $8$ at $x=3$. **Final answer:** The function $f(x) = x^2 - 2x + 5$ has vertex at $(1,4)$ and on $[0,3]$ the values go from $5$ to $8$ with a minimum at $4$.