Find the range of the function: $$ y=x^{2}-3 x $$

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Write the function as $$ y = x^2 - 3x $$.
Express $$ y = x^2-3x $$ in the form $$ y = \left(x-\frac{3}{2}\right)^2 - \frac{9}{4} $$. This is done by noting $$ x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 $$.
Since $$ \left(x-\frac{3}{2}\right)^2 $$ is always nonnegative, its smallest value is $$ 0 $$. Hence, the minimum value of $$ y $$ is $$ -\frac{9}{4} $$.
The range is all $$ y $$ values greater than or equal to $$ -\frac{9}{4} $$.
The range of the function is $$ \left[-\frac{9}{4}, \infty\right) $$.

Find the range of the function:

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