Quadratic Graph 0A1280

1. **State the problem:** We need to analyze and understand the graph of the quadratic equation $$y = x^2 + 2x - 2$$. 2. **Recall the standard form:** A quadratic equation is generally written as $$y = ax^2 + bx + c$$, where $$a$$, $$b$$, and $$c$$ are constants. 3. **Identify coefficients:** Here, $$a = 1$$, $$b = 2$$, and $$c = -2$$. 4. **Find the vertex:** The vertex of a parabola given by $$y = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. Calculate: $$x = -\frac{2}{2 \times 1} = -1$$ Find $$y$$ at $$x = -1$$: $$y = (-1)^2 + 2(-1) - 2 = 1 - 2 - 2 = -3$$ So, the vertex is $$(-1, -3)$$. 5. **Determine the axis of symmetry:** The axis of symmetry is the vertical line through the vertex: $$x = -1$$. 6. **Find the y-intercept:** Set $$x = 0$$: $$y = 0^2 + 2(0) - 2 = -2$$ So, the y-intercept is at $$(0, -2)$$. 7. **Find the x-intercepts (roots):** Solve $$x^2 + 2x - 2 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}$$ So, the roots are: $$x = -1 + \sqrt{3}$$ and $$x = -1 - \sqrt{3}$$. 8. **Shape and direction:** Since $$a = 1 > 0$$, the parabola opens upwards. **Summary:** The parabola has vertex at $$(-1, -3)$$, axis of symmetry $$x = -1$$, y-intercept at $$(0, -2)$$, and x-intercepts at $$-1 \pm \sqrt{3}$$. It opens upwards. **Final answer:** The graph is a parabola opening upwards with the described features.