1. **State the problem:** We are given four quadratic equations and need to find their horizontal intercepts and the x-coordinate of their vertex without graphing. 2. **Recall the formulas:** - Horizontal intercepts (roots) of $y = ax^2 + bx + c$ are found by solving $ax^2 + bx + c = 0$. - For equations with $c=0$, roots are $x=0$ and $x=-\frac{b}{a}$. - The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$. 3. **Apply to each equation:** **Equation 1: $y = x^2 + 6x$** - Here, $a=1$, $b=6$, $c=0$. - Roots: $x=0$ and $x=-\frac{6}{1} = -6$. - Vertex x-coordinate: $x = -\frac{6}{2 \times 1} = -3$. **Equation 2: $y = x^2 - 10x$** - $a=1$, $b=-10$, $c=0$. - Roots: $x=0$ and $x=-\frac{-10}{1} = 10$. - Vertex x-coordinate: $x = -\frac{-10}{2 \times 1} = 5$. **Equation 3: $y = -x^2 + 50x$** - $a=-1$, $b=50$, $c=0$. - Roots: $x=0$ and $x=-\frac{50}{-1} = 50$. - Vertex x-coordinate: $x = -\frac{50}{2 \times -1} = 25$. **Equation 4: $y = -x^2 - 36x$** - $a=-1$, $b=-36$, $c=0$. - Roots: $x=0$ and $x=-\frac{-36}{-1} = -36$. - Vertex x-coordinate: $x = -\frac{-36}{2 \times -1} = -18$. 4. **Summary table:** | Equation | Horizontal Intercept 1 $(x,y)$ | Horizontal Intercept 2 $(x,y)$ | Vertex x-coordinate | |---|---|---|---| | $y = x^2 + 6x$ | $(0,0)$ | $(-6,0)$ | $-3$ | | $y = x^2 - 10x$ | $(0,0)$ | $(10,0)$ | $5$ | | $y = -x^2 + 50x$ | $(0,0)$ | $(50,0)$ | $25$ | | $y = -x^2 - 36x$ | $(0,0)$ | $(-36,0)$ | $-18$ | This completes the table without graphing the equations.