Find the range of $y = x^{2} + 3 x + 2$

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Solution

By making perfect square or pure quadratic, we get
$y = {(x + \frac{3}{2})}^{2} + 2 - \frac{9}{4}$
$= {(x + \frac{3}{2})}^{2} + \frac{8 - 9}{4}$
$= {(x + \frac{3}{2})}^{2} - \frac{1}{4}$ as ${(x + \frac{3}{2})}^{2} \geq 0$ for all $x \in R$
Thus, ${(x + \frac{3}{2})}^{2} - \frac{1}{4} \geq \frac{- 1}{4}$
Hence range of $f (x) \in [\frac{- 1}{4}, \infty)$

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