Given y = x2 + ax + b, axis of symmetry x = -2, and passing through (-1,-5). Find a and b. Given y = (x+k)2 + 3x − 2y, find k if the minimum value is 5/3.

The given quadratic equation is $y = x^2 + ax + b$.
The axis of symmetry for a quadratic $y = Ax^2 + Bx + C$ is given by the formula $x = -\frac{B}{2A}$.
In this equation, $A=1$ and $B=a$.
So, the axis of symmetry is $x = -\frac{a}{2}$.
We are given that the axis of symmetry is $x = -2$.

$$-\frac{a}{2} = -2$$ $$a = 4$$

The graph passes through the point $(-1, -5)$. Substitute $x=-1$, $y=-5$, and $a=4$ into the equation $y = x^2 + ax + b$:

$$-5 = (-1)^2 + 4(-1) + b$$ $$-5 = 1 - 4 + b$$ $$-5 = -3 + b$$ $$b = -5 + 3$$ $$b = -2$$

The values are $a=4$ and $b=-2$.

The given equation is $y = (x+k)^2 + 3x - 2y$.
First, rearrange the equation into the standard quadratic form $y = Ax^2 + Bx + C$:

$$y = x^2 + 2kx + k^2 + 3x - 2y$$

Add $2y$ to both sides:

$$3y = x^2 + (2k+3)x + k^2$$

Divide by 3:

$$y = \frac{1}{3}x^2 + \frac{2k+3}{3}x + \frac{k^2}{3}$$

For a quadratic equation $y = Ax^2 + Bx + C$, the minimum value (since $A = \frac{1}{3} > 0$) occurs at the vertex and is given by $C - \frac{B^2}{4A}$.
In this equation, $A = \frac{1}{3}$, $B = \frac{2k+3}{3}$, and $C = \frac{k^2}{3}$.
The minimum value is:

$$\frac{k^2}{3} - \frac{\left(\frac{2k+3}{3}\right)^2}{4\left(\frac{1}{3}\right)}$$ $$= \frac{k^2}{3} - \frac{\frac{(2k+3)^2}{9}}{\frac{4}{3}}$$ $$= \frac{k^2}{3} - \frac{(2k+3)^2}{9} \cdot \frac{3}{4}$$ $$= \frac{k^2}{3} - \frac{(2k+3)^2}{12}$$

We are given that the minimum value is $-4$.

$$\frac{k^2}{3} - \frac{(2k+3)^2}{12} = -4$$

Multiply the entire equation by 12 to eliminate the denominators:

$$4k^2 - (2k+3)^2 = -48$$

Expand $(2k+3)^2$:

$$4k^2 - (4k^2 + 12k + 9) = -48$$ $$4k^2 - 4k^2 - 12k - 9 = -48$$ $$-12k - 9 = -48$$

Add 9 to both sides:

$$-12k = -48 + 9$$ $$-12k = -39$$

Divide by $-12$:

$$k = \frac{-39}{-12}$$ $$k = \frac{13}{4}$$

The final answers are $\boxed{a=4, b=-2}$ and $\boxed{k=\frac{13}{4}}$.