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

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Answer 1 · 2015-09-04T06:34:28

Maximum is + $oo$ and Minimum is 12

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]}

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