1. The problem asks to find the vertex of each quadratic function using the formula for the x-coordinate of the vertex: $$x = -\frac{b}{2a}$$ where the quadratic is in the form $$y = ax^2 + bx + c$$. 2. For each quadratic, we will find the x-coordinate of the vertex using the formula, then substitute it back into the function to find the y-coordinate. 3. (a) For $$y = x^2 + 6x + 2$$, here $$a=1$$ and $$b=6$$. Calculate $$x$$: $$x = -\frac{6}{2 \times 1} = -\frac{6}{2} = -3$$ Substitute $$x = -3$$ into the function: $$y = (-3)^2 + 6(-3) + 2 = 9 - 18 + 2 = -7$$ So the vertex is $$(-3, -7)$$. 4. (b) For $$y = 3x^2 - 12x + 5$$, here $$a=3$$ and $$b=-12$$. Calculate $$x$$: $$x = -\frac{-12}{2 \times 3} = \frac{12}{6} = 2$$ Substitute $$x = 2$$ into the function: $$y = 3(2)^2 - 12(2) + 5 = 3(4) - 24 + 5 = 12 - 24 + 5 = -7$$ So the vertex is $$(2, -7)$$. 5. (c) For $$y = -x^2 + 8x - 11$$, here $$a=-1$$ and $$b=8$$. Calculate $$x$$: $$x = -\frac{8}{2 \times (-1)} = -\frac{8}{-2} = 4$$ Substitute $$x = 4$$ into the function: $$y = -(4)^2 + 8(4) - 11 = -16 + 32 - 11 = 5$$ So the vertex is $$(4, 5)$$. Final answers: - (a) Vertex: $$(-3, -7)$$ - (b) Vertex: $$(2, -7)$$ - (c) Vertex: $$(4, 5)$$