Question

How do you find the maximum value of `y=x^2+2x+3`?

Answer 1

There is no maximum value.

Explanation:

1) The graph is an upward opening parabola; there is no maximum.

2) `lim_(xrarroo)(x^2+2x+3) = oo`, so there is no maximum.

Using `y = f(x)`,

3) `f'(x) = 2x+2 = 0` at `x=-1`. Therefore, `-1` is the only critical number and the first or second derivative test will show that `f(-1)` is a minimum, not a maximum.