Quadratic Functions and Equations

Half of $-6$ is $-3$, and $(x-3)^2 = x^2 - 6x + 9$. Since the constant is 11, we need $11 - 9 = 2$ more.

$b^2 - 4ac = 9 - 4(2)(-5) = 9 + 40 = 49$. It is positive, so there are two distinct real roots, and because 49 is a perfect square they are rational: $(2x+5)(x-1) = 0$ gives $x = 1$ and $x = -\tfrac52$.

$3 + 4x - x^2 = -(x^2 - 4x) + 3 = -\big[(x-2)^2 - 4\big] + 3 = -(x-2)^2 + 7$. The bracket is never negative, so the largest value is 7, reached when $x = 2$.

No real roots means $\Delta < 0$: $(k-1)^2 - 16 < 0$, so $(k-1)^2 < 16$, giving $-4 < k - 1 < 4$ and therefore $-3 < k < 5$.

$\dfrac1\alpha + \dfrac1\beta = \dfrac{\beta + \alpha}{\alpha\beta} = \dfrac{7}{2}$, since $\alpha+\beta = 7$ and $\alpha\beta = 2$.

Setting them equal: $x^2 + 3x + 4 = 2x + 1$, so $x^2 + x + 3 = 0$. Its discriminant is $1 - 12 = -11$, which is negative, so this equation has no real solutions and the graphs do not intersect.