1. **State the problem:** We need to plot the parabola $Q$ given by the equation $$y = \frac{(x+6)(x+2)}{4}$$ and understand its shape and key points. 2. **Formula and explanation:** The function is a quadratic function in factored form. The general form of a quadratic is $$y = a(x-r_1)(x-r_2)$$ where $r_1$ and $r_2$ are roots (x-intercepts). Here, $a = \frac{1}{4}$, $r_1 = -6$, and $r_2 = -2$. 3. **Find the roots:** Set $y=0$: $$0 = \frac{(x+6)(x+2)}{4} \implies (x+6)(x+2) = 0$$ So, $x = -6$ or $x = -2$ are the roots. 4. **Find the vertex:** The vertex $x$-coordinate is the midpoint of the roots: $$x_v = \frac{-6 + (-2)}{2} = \frac{-8}{2} = -4$$ 5. **Calculate the vertex $y$-value:** $$y_v = \frac{(-4+6)(-4+2)}{4} = \frac{(2)(-2)}{4} = \frac{-4}{4} = -1$$ 6. **Plot points of interest:** - At $x=-6$, $y=0$ - At $x=-2$, $y=0$ - At $x=-4$, $y=-1$ (vertex, minimum point) 7. **Shape and direction:** Since $a=\frac{1}{4} > 0$, the parabola opens upwards. 8. **Summary:** The parabola crosses the x-axis at $-6$ and $-2$, has a minimum vertex at $(-4, -1)$, and opens upward with a relatively wide shape due to the coefficient $\frac{1}{4}$. **Final answer:** The parabola $Q$ is $$y = \frac{(x+6)(x+2)}{4}$$ with roots at $x=-6$ and $x=-2$, vertex at $(-4,-1)$, opening upwards.