1. **State the problem:** We analyze the quadratic function $$f(x) = x^2 + 2x - 3$$ to find its domain, range, vertex, axis of symmetry, intercepts, and key attributes. 2. **Formula and rules:** A quadratic function is generally $$f(x) = ax^2 + bx + c$$ with vertex at $$x = -\frac{b}{2a}$$ and vertex value $$f\left(-\frac{b}{2a}\right)$$. 3. **Find vertex:** Here, $$a=1$$, $$b=2$$, $$c=-3$$. Calculate $$x$$-coordinate of vertex: $$x = -\frac{b}{2a} = -\frac{2}{2 \times 1} = -1$$ Calculate $$y$$-coordinate: $$f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$$ So vertex is $$(-1, -4)$$. 4. **Axis of symmetry:** The vertical line through the vertex: $$x = -1$$ 5. **Find y-intercept:** Set $$x=0$$: $$f(0) = 0 + 0 - 3 = -3$$ So y-intercept is $$(0, -3)$$. 6. **Find x-intercepts (roots):** Solve $$x^2 + 2x - 3 = 0$$. Factor: $$x^2 + 2x - 3 = (x + 3)(x - 1) = 0$$ So roots are: $$x = -3$$ and $$x = 1$$ 7. **Domain:** Quadratic functions have domain all real numbers: $$\text{Domain} = (-\infty, \infty)$$ 8. **Range:** Since $$a=1 > 0$$, parabola opens upward, so minimum value is at vertex $$y = -4$$. Thus, $$\text{Range} = [-4, \infty)$$ 9. **Summary:** - Vertex: $$(-1, -4)$$ - Axis of symmetry: $$x = -1$$ - x-intercepts: $$(-3, 0), (1, 0)$$ - y-intercept: $$(0, -3)$$ - Domain: all real numbers - Range: $$y \geq -4$$