QuestionFind the max or min of the function: $f(x)=-x^{2}-3x+3$.

Studdy Solution
Therefore, the vertex of the parabola is $(\frac{3}{2}, -\frac{15}{4})$. Since the coefficient of $x^{2}$ is negative, the parabola opens downwards, and the vertex is the maximum point.
The maximum value of the function $f(x)=-x^{2}-3x+3$ is $-\frac{15}{4}$.