1. **State the problem:** We are given the quadratic function $$y = x^2 + 2x$$ and need to analyze it. 2. **Recall the standard form:** A quadratic function 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 $c = 0$. 4. **Find the vertex:** The vertex of a parabola given by $$y = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. Calculate: $$x = -\frac{2}{2 \times 1} = -\frac{2}{2} = -1$$ 5. **Find the y-coordinate of the vertex:** Substitute $x = -1$ into the function: $$y = (-1)^2 + 2(-1) = 1 - 2 = -1$$ So the vertex is at $$(-1, -1)$$. 6. **Find the y-intercept:** Set $x=0$: $$y = 0^2 + 2 \times 0 = 0$$ So the y-intercept is at $(0,0)$. 7. **Find the x-intercepts:** Set $y=0$ and solve for $x$: $$0 = x^2 + 2x$$ Factor: $$0 = x(x + 2)$$ So, $$x = 0 \quad \text{or} \quad x = -2$$ 8. **Summary:** - Vertex: $(-1, -1)$ - Y-intercept: $(0,0)$ - X-intercepts: $(0,0)$ and $(-2,0)$ This parabola opens upwards since $a=1 > 0$. Final answer: The vertex is at $(-1, -1)$, the parabola crosses the x-axis at $x=0$ and $x=-2$, and the y-axis at $y=0$.