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

##### 1 Answer

Maximum is +

#### Explanation:

Write the function in the vertex form,

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

This represents a vertical parabola opening upwards with its vertex at

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