1. **State the problem:** We want to find values of $m$ such that the vertices of the parabolas $$y=-x^2-4mx+m$$ and $$y=x^2-2mx-1$$ both lie on the same side of the X-axis, meaning both their $y$-coordinates are either positive or negative. 2. **Find the vertex of the first parabola:** The general form is $$y = ax^2 + bx + c$$ with $a=-1$, $b=-4m$, and $c=m$. Vertex $x$-coordinate: $$x_1 = -\frac{b}{2a} = -\frac{-4m}{2(-1)} = -\frac{-4m}{-2} = -2m$$ Calculate $y_1$: $$y_1 = -(-2m)^2 -4m(-2m) + m = -4m^2 + 8m^2 + m = 4m^2 + m$$ 3. **Find the vertex of the second parabola:** Here $a=1$, $b=-2m$, $c=-1$. Vertex $x$-coordinate: $$x_2 = -\frac{b}{2a} = -\frac{-2m}{2} = m$$ Calculate $y_2$: $$y_2 = (m)^2 - 2m(m) - 1 = m^2 - 2m^2 - 1 = -m^2 - 1$$ 4. **Analyze sign conditions:** - Vertex $y$-coordinate of first parabola: $$y_1 = 4m^2 + m$$ - Vertex $y$-coordinate of second parabola: $$y_2 = -m^2 - 1$$ Since $$y_2 = -m^2 -1 < 0$$ for all real $m$ (because $-m^2 \\leq 0$ and $-1$ makes it strictly negative), the second vertex is always below the X-axis. 5. **Both vertices on the same side means both $y_1$ and $y_2$ are negative:** We require: $$4m^2 + m < 0$$ 6. **Solve inequality:** $$4m^2 + m < 0$$ Factor out $m$: $$m(4m + 1) < 0$$ The product is negative when the two factors have opposite signs: - Case 1: $m < 0$ and $4m + 1 > 0$\Rightarrow $ $m > -\frac{1}{4}$ - Case 2: $m > 0$ and $4m + 1 < 0$ (impossible since $4m + 1 > 0$ if $m > 0$) So the solution interval is: $$-\frac{1}{4} < m < 0$$ 7. **Check the multiple-choice points provided:** Among the points $-\frac{1}{2}$, $-\frac{1}{4}$, $\frac{1}{2}$, and $\frac{1}{4}$, values inside $(-\frac{1}{4}, 0)$ include no exact options but the nearest point just inside is between $-\frac{1}{4}$ and $0$. Thus, the vertex values lie on the same side of the X-axis (below) for: $$m \in (-\frac{1}{4}, 0)$$ **Final answer:** $m$ values between $-\frac{1}{4}$ and $0$.