1. **State the problem:** We are given the quadratic function $f(x) = x^2 + 2x - 3$ and want to analyze it. 2. **Formula and rules:** A quadratic function is generally written as $f(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. Here, $a=1$, $b=2$, and $c=-3$. 3. **Find the roots (x-intercepts):** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Calculate the discriminant: $$\Delta = b^2 - 4ac = 2^2 - 4 \times 1 \times (-3) = 4 + 12 = 16$$ Since $\Delta > 0$, there are two real roots. 4. **Calculate roots:** $$x = \frac{-2 \pm \sqrt{16}}{2 \times 1} = \frac{-2 \pm 4}{2}$$ So, $$x_1 = \frac{-2 + 4}{2} = 1$$ $$x_2 = \frac{-2 - 4}{2} = -3$$ 5. **Find the vertex (extremum):** The vertex $x$-coordinate is given by $$x = -\frac{b}{2a} = -\frac{2}{2 \times 1} = -1$$ Calculate $f(-1)$: $$f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$$ So the vertex is at $(-1, -4)$, which is a minimum point since $a > 0$. 6. **Summary:** - Roots (x-intercepts): $x=1$ and $x=-3$ - Vertex (minimum): $(-1, -4)$ - The parabola opens upwards because $a=1 > 0$. Final answer: The function $f(x) = x^2 + 2x - 3$ has roots at $x=1$ and $x=-3$, and a minimum vertex at $(-1, -4)$.