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Find the smallest possible value of (n^2 + 3n + 19)^2, if n is an integer.

Suchen Sie den kleinstmöglichen Wert von (n ^ 2 + 3n + 19) ^ 2, wenn n eine ganze Zahl ist.

Hello Guest!

$f(n)=(n^2 + 3n + 19)^2\\ f'(n)=2(n^2 + 3n + 19)(2n+3)=0\\ 2n+3=0\\ n_1=-1.5\\ f(-1)=\color{blue}289\\ f(-1.5=280.5625\ minimum\\ f(-2)=\color{blue}289$

$n^2+3n+19=0\\ n=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}\\ n=-\frac{3}{2}\pm \sqrt{2.25-19}\\ n_{2,3}\ are\ complex\ numbers.$

$The\ smallest\ value\ of\ f(n)\ is\ 289\ |n\in \mathbb{Z}.$

  ! 

asinus

you do $\frac{-b}{2a}$.

Then plug in and solve