1. **State the problem:** We are given the quadratic function $y = x^2 + 2x - 3$ and want to analyze it. 2. **Formula and rules:** The general form of a quadratic function is $y = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. 3. **Find the roots (x-intercepts):** Set $y=0$ and solve for $x$: $$x^2 + 2x - 3 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=2$, $c=-3$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 2^2 - 4 \times 1 \times (-3) = 4 + 12 = 16$$ 5. **Calculate the roots:** $$x = \frac{-2 \pm \sqrt{16}}{2 \times 1} = \frac{-2 \pm 4}{2}$$ 6. **Evaluate each root:** - For $+$ sign: $$x = \frac{-2 + 4}{2} = \frac{2}{2} = 1$$ - For $-$ sign: $$x = \frac{-2 - 4}{2} = \frac{-6}{2} = -3$$ 7. **Vertex of the parabola:** The vertex $x$-coordinate is given by: $$x = -\frac{b}{2a} = -\frac{2}{2 \times 1} = -1$$ 8. **Calculate the vertex $y$-coordinate:** $$y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$$ 9. **Summary:** - Roots (x-intercepts) are $x=1$ and $x=-3$. - Vertex is at $(-1, -4)$. - Since $a=1 > 0$, the parabola opens upwards. Final answer: The quadratic $y = x^2 + 2x - 3$ has roots at $x=1$ and $x=-3$, and its vertex is at $(-1, -4)$.