Setting the derivative f'(-1) = 0 gives k = 1 for differentiability.

f'(x) = 2x - 2. Thus, f'(1) = 2(1) - 2 = 0.

As a polynomial, f(x) is differentiable everywhere, including at x = 2.

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1, which is the minimum value.

The vertex is at x = -b/(2a) = 4/2 = 2. f(2) = 2^2 - 4(2) + 5 = 1, so the vertex is (2, 1).

The x-coordinate of the vertex is given by x = -b/(2a) = 6/(2*1) = 3.

As a polynomial, f(x) is differentiable everywhere, hence no points of non-differentiability.

The function is a polynomial and is differentiable everywhere, hence there are no points where it is not differentiable.

The function is a polynomial and is differentiable everywhere, so there is no such x.

f(x) is a polynomial and is differentiable everywhere. The x-coordinate can be any real number.

First, find f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 3)(x - 1) = 0, so critical points are x = 1 and x = 3.

f'(x) = 3x^2 - 12x + 9. The critical points are x = 1 and x = 3. The function is increasing on (1, 3) and (3, ∞).

f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f(2) = 1 is a local minimum.

Find f''(x) = 12x^2 - 16. Setting f''(x) = 0 gives x^2 = 4, so x = ±2. f(2) = 0, thus the inflection point is (2, 0).

f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x(x^2 - 4) = 0, so x = -2, 0, 2. Test intervals: f' is positive in (-2, ∞).

Setting the left limit (0 + 1 = 1) equal to the right limit (3 - 0 = 3), we find b = 1.

By definition, f(1) = 3, as given in the piecewise function.

Setting f(1-) = f(1+) and f'(1-) = f'(1+) gives k = 2 for differentiability.

For continuity, f(3) must equal the limit as x approaches 3, which is 9.

The function is not differentiable at x = -3 and x = 2, but the first point of interest is -3.

The distance between the foci of the hyperbola is 2c, where c = √(a^2 + b^2) = √(25 + 16) = √41, so the distance is 2√41.

The slopes can be found by solving the quadratic equation derived from the given equation.

The product of the slopes of the lines can be found from the equation, which gives -1.

The sum of the slopes can be found using the relationship between the coefficients of the quadratic.

The condition for parallel lines is that the determinant of the coefficients must equal zero.

Factoring the equation gives the slopes as m1 = 1 and m2 = 3.

The lines are at an angle of 45 degrees as the determinant of the coefficients gives a non-zero value.

The sum of the slopes is given by - (coefficient of xy)/(coefficient of x^2) = -6/5.

The sum of the slopes of the lines is given by -b/a, which is -5/6.

The slopes can be found by solving the quadratic equation formed by the coefficients of x^2, xy, and y^2.