1. **State the problem:** We are given two quadratic functions: - $y = -2(x - 1)^2 + 3$ - $y = (x - 5)^2 + 6$ We need to find the vertex, axis of symmetry, additional points, number and type of solutions, max or min, domain, and range for each. 2. **Recall the vertex form of a quadratic:** $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. 3. **For the first function $y = -2(x - 1)^2 + 3$:** - Vertex: $(1, 3)$ because $h=1$, $k=3$. - Axis of symmetry: vertical line $x = 1$. - Since $a = -2 < 0$, the parabola opens downward, so the vertex is a maximum point. - Domain: all real numbers, $(-\infty, \infty)$. - Range: since max is at $y=3$ and parabola opens down, range is $(-\infty, 3]$. 4. **Find second and third points:** Choose $x=0$ and $x=2$ (one unit left and right of vertex): - At $x=0$: $$y = -2(0 - 1)^2 + 3 = -2(1)^2 + 3 = -2 + 3 = 1$$ - At $x=2$: $$y = -2(2 - 1)^2 + 3 = -2(1)^2 + 3 = -2 + 3 = 1$$ Points: $(0,1)$ and $(2,1)$. 5. **Number and type of solutions:** Set $y=0$ to find roots: $$0 = -2(x - 1)^2 + 3$$ $$-2(x - 1)^2 = -3$$ $$\cancel{-2}(x - 1)^2 = \cancel{-3} \Rightarrow (x - 1)^2 = \frac{3}{2}$$ $$x - 1 = \pm \sqrt{\frac{3}{2}}$$ $$x = 1 \pm \sqrt{\frac{3}{2}}$$ Two distinct real solutions. 6. **For the second function $y = (x - 5)^2 + 6$:** - Vertex: $(5, 6)$. - Axis of symmetry: $x = 5$. - Since $a=1 > 0$, parabola opens upward, vertex is minimum. - Domain: $(-\infty, \infty)$. - Range: $[6, \infty)$. 7. **Graph shapes:** - First parabola opens downward with vertex at $(1,3)$. - Second parabola opens upward with vertex at $(5,6)$. Final answers: - First function vertex: $(1,3)$, axis $x=1$, max at $y=3$, domain $(-\infty, \infty)$, range $(-\infty, 3]$, two real solutions. - Second function vertex: $(5,6)$, axis $x=5$, min at $y=6$, domain $(-\infty, \infty)$, range $[6, \infty)$.