1. **State the problem:** Write the quadratic expression $x^2 - 13x + 3$ in the form $(x + a)^2 + b$, where $a$ and $b$ are constants. 2. **Recall the formula:** To complete the square for a quadratic expression $x^2 + bx + c$, we use the identity: $$x^2 + bx + c = (x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2 + c$$ This helps rewrite the quadratic in vertex form. 3. **Identify coefficients:** Here, $b = -13$ and $c = 3$. 4. **Calculate $a$:** $$a = \frac{b}{2} = \frac{-13}{2} = -\frac{13}{2}$$ 5. **Rewrite the expression:** $$x^2 - 13x + 3 = \left(x - \frac{13}{2}\right)^2 - \left(\frac{13}{2}\right)^2 + 3$$ 6. **Simplify the constants:** $$- \left(\frac{13}{2}\right)^2 + 3 = - \frac{169}{4} + 3 = - \frac{169}{4} + \frac{12}{4} = - \frac{157}{4}$$ 7. **Final form:** $$\boxed{\left(x - \frac{13}{2}\right)^2 - \frac{157}{4}}$$ This is the expression in the form $(x + a)^2 + b$ with $a = -\frac{13}{2}$ and $b = -\frac{157}{4}$. The vertex of the parabola is at $\left(\frac{13}{2}, -\frac{157}{4}\right)$, confirming the bottom-left location of the vertex as described.