1. **State the problem:** We are given the quadratic function $y = x^2 - 6x + 10$ and the equation $1 = x - 3$. We want to analyze the function and solve for $x$ in the equation. 2. **Solve the linear equation:** From $1 = x - 3$, add 3 to both sides to isolate $x$: $$x = 1 + 3 = 4$$ 3. **Evaluate the quadratic function at $x=4$:** Substitute $x=4$ into $y = x^2 - 6x + 10$: $$y = 4^2 - 6(4) + 10 = 16 - 24 + 10 = 2$$ 4. **Summary:** The value of $y$ when $x=4$ is $2$. This means the point $(4, 2)$ lies on the parabola defined by the quadratic function. 5. **Additional notes:** The quadratic function $y = x^2 - 6x + 10$ can be rewritten in vertex form by completing the square: $$y = (x^2 - 6x) + 10 = (x^2 - 6x + 9) + 10 - 9 = (x - 3)^2 + 1$$ This shows the vertex is at $(3, 1)$, which is the minimum point of the parabola. Final answer: When $x=4$, $y=2$.