1. **State the problem:** Find the point where the quadratic function $y = x^2 + 2x + 1$ touches the x-axis. 2. **Recall the formula:** The quadratic touches the x-axis where $y=0$, so solve the equation: $$x^2 + 2x + 1 = 0$$ 3. **Factor the quadratic:** $$x^2 + 2x + 1 = (x+1)^2$$ 4. **Set the factor equal to zero:** $$(x+1)^2 = 0$$ 5. **Solve for $x$:** $$x + 1 = 0 \\ x = -1$$ 6. **Interpretation:** The quadratic touches the x-axis at $x = -1$, and since it is a perfect square, it only touches (tangent) at this point. 7. **Find the corresponding $y$ value:** $$y = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0$$ **Final answer:** The quadratic touches the x-axis at the point $(-1, 0)$.