Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = ln(x2 + 3x + 4), [−2, 2]

The vertex (minimum) of the quadratic ax² +bx +c is located at x=-b/(2a). This means the minimum value of f(x) will be found at x = -3/(2*1) = -1.5.

Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.

On the given interval, ...
  the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
  the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057

Answer d; a supplementary angle = 180°, so if that angle = 150°, y would = 30°.