For what real value of b is the expression \(\frac{1}{2}b^2 + 5b - 3\)minimized?
+23251 One way to do this is to find the vertex of this quadratic: y = 0.5b2 + 5b - 3
Move the 3 to the other side: y + 3 = 0.5b2 + 5b
Factor out the 0.5: y + 3 = 0.5(b2 + 10b)
Complete the square: y + 3 + 12.5 = 0.5(b2 + 10b + 25)
Factor and simplify: y + 15.5 = 0.5(b + 5)2
The vertex is (-5, -15.5) which means that the minimum is -15.5 and this occurs at -5.
