1. The problem is to find the vertex of each quadratic function given. 2. The vertex form of a quadratic function is $$y = a(x-h)^2 + k$$ where $(h,k)$ is the vertex. 3. For functions in the form $$y = ax^2 + bx + c$$, the vertex $x$-coordinate is given by $$h = -\frac{b}{2a}$$ and the $y$-coordinate is $$k = y(h)$$. 4. Since all given functions have $b=0$, the vertex $x$-coordinate is always $$h = -\frac{0}{2a} = 0$$. 5. Now, find the $y$-coordinate by substituting $x=0$ into each function: (1) $$y = x^2$$ $$y(0) = 0^2 = 0$$ Vertex: $(0,0)$ (correcting the given vertex) (2) $$y = x^2 + 1$$ $$y(0) = 0^2 + 1 = 1$$ Vertex: $(0,1)$ (3) $$y = x^2 + 3$$ $$y(0) = 0^2 + 3 = 3$$ Vertex: $(0,3)$ (4) $$y = 3x^2$$ $$y(0) = 3 \times 0^2 = 0$$ Vertex: $(0,0)$ (5) $$y = 3x^2 - 1$$ $$y(0) = 3 \times 0^2 - 1 = -1$$ Vertex: $(0,-1)$ (6) $$y = 3x^2 + 2$$ $$y(0) = 3 \times 0^2 + 2 = 2$$ Vertex: $(0,2)$ (7) $$y = -x^2 + 3$$ $$y(0) = -0^2 + 3 = 3$$ Vertex: $(0,3)$ (8) $$y = -3x^2 - 5$$ $$y(0) = -3 \times 0^2 - 5 = -5$$ Vertex: $(0,-5)$ Final answers: (1) $(0,0)$ (2) $(0,1)$ (3) $(0,3)$ (4) $(0,0)$ (5) $(0,-1)$ (6) $(0,2)$ (7) $(0,3)$ (8) $(0,-5)$