1. **Problem:** Sketch the graph of the function $y = -2(x - 5)^2 + 8$. 2. **Formula and rules:** This is a quadratic function in vertex form $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex and $a$ determines the direction and width of the parabola. 3. **Identify vertex and direction:** Here, $a = -2$, $h = 5$, and $k = 8$. The vertex is at $(5,8)$ and since $a < 0$, the parabola opens downward. 4. **Plot vertex and shape:** The graph is a downward-opening parabola with vertex at $(5,8)$. 5. **Find intercepts:** To find the $x$-intercepts, set $y=0$: $$0 = -2(x - 5)^2 + 8$$ $$-2(x - 5)^2 = -8$$ $$\cancel{-2}(x - 5)^2 = \cancel{-8}4$$ $$ (x - 5)^2 = 4$$ $$ x - 5 = \pm 2$$ $$ x = 5 \pm 2$$ So, $x=3$ or $x=7$. 6. **Find $y$-intercept:** Set $x=0$: $$y = -2(0 - 5)^2 + 8 = -2(25) + 8 = -50 + 8 = -42$$ 7. **Summary:** The parabola has vertex $(5,8)$, $x$-intercepts at $(3,0)$ and $(7,0)$, and $y$-intercept at $(0,-42)$. --- 1. **Problem:** Sketch the graph of the function $y = 3(x - 1)(x + 3)$. 2. **Formula and rules:** This is a quadratic function in factored form $y = a(x - r_1)(x - r_2)$, where $r_1$ and $r_2$ are roots. 3. **Identify roots and direction:** Here, $a=3$, roots are $x=1$ and $x=-3$. Since $a>0$, the parabola opens upward. 4. **Find vertex:** The vertex lies midway between roots: $$x_v = \frac{1 + (-3)}{2} = \frac{-2}{2} = -1$$ 5. **Calculate $y$ at vertex:** $$y_v = 3(-1 - 1)(-1 + 3) = 3(-2)(2) = -12$$ 6. **Find $y$-intercept:** Set $x=0$: $$y = 3(0 - 1)(0 + 3) = 3(-1)(3) = -9$$ 7. **Summary:** The parabola opens upward with roots at $(1,0)$ and $(-3,0)$, vertex at $(-1,-12)$, and $y$-intercept at $(0,-9)$. --- **Final answers:** - For $y = -2(x - 5)^2 + 8$, vertex at $(5,8)$, $x$-intercepts at $(3,0)$ and $(7,0)$, $y$-intercept at $(0,-42)$. - For $y = 3(x - 1)(x + 3)$, roots at $(1,0)$ and $(-3,0)$, vertex at $(-1,-12)$, $y$-intercept at $(0,-9)$.