Question: 6. g(x) = x² + 11x
1. **State the problem:** We are given the function $$g(x) = x^2 + 11x$$ and want to analyze its properties such as vertex and x-intercepts.
2. **Formula and rules:** This is a quadratic function in the form $$g(x) = ax^2 + bx + c$$ where $$a=1$$, $$b=11$$, and $$c=0$$.
- The vertex of a parabola $$y = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$.
- The x-intercepts are found by solving $$g(x) = 0$$.
3. **Find the vertex:**
$$x = -\frac{b}{2a} = -\frac{11}{2 \times 1} = -\frac{11}{2}$$
4. **Find the y-coordinate of the vertex:**
$$g\left(-\frac{11}{2}\right) = \left(-\frac{11}{2}\right)^2 + 11 \times \left(-\frac{11}{2}\right) = \frac{121}{4} - \frac{121}{2} = \frac{121}{4} - \frac{242}{4} = -\frac{121}{4}$$
5. **Find the x-intercepts:** Solve $$x^2 + 11x = 0$$
Factor:
$$x(x + 11) = 0$$
Set each factor to zero:
$$x = 0$$ or $$x + 11 = 0 \Rightarrow x = -11$$
6. **Summary:**
- Vertex at $$\left(-\frac{11}{2}, -\frac{121}{4}\right)$$
- x-intercepts at $$x=0$$ and $$x=-11$$
This matches the description of an upward-opening parabola with vertex at $$x = -\frac{11}{2}$$ and x-intercepts at $$0$$ and $$-11$$.