Question

How do you find the maximum or minimum of `y=x^2+4x+16`?

Answer 1

Maximum is +`oo` and Minimum is 12

Explanation:

Write the function in the vertex form,

`y= (x+2)^2 +12`

This represents a vertical parabola opening upwards with its vertex at `(-2,12)`. The minimum value of `y` is 12 for `x=-2`

graph{x^2 + 4x + 16 [-40.76, 41.44, -7.73, 33.4]}