Sendo f (x)= 3x^2+10x-3 , determine f(2) f(0) f(-2) f(1/2)^=elevado ​

\[\eqalign{ f\left( 2 \right) = 3 \cdot {2^2} + 10 \cdot 2 - 3 \cr = 3 \cdot 4 + 20 - 3 \cr = 12 + 20 - 3 \cr = 29 \cr \cr \Rightarrow \boxed{f\left( 2 \right) = 29} \cr \cr f\left( 0 \right) = 3 \cdot {0^2} + 10 \cdot 0 - 3 \cr = 3 \cdot 0 + 0 - 3 \cr = 0 + 0 - 3 \cr = - 3 \cr \cr \Rightarrow \boxed{f\left( 0 \right) = - 3} \cr \cr f\left( { - 2} \right) = 3 \cdot {\left( { - 2} \right)^2} + 10 \cdot \left( { - 2} \right) - 3 \cr = 3 \cdot 4 - 20 - 3 \cr = 12 - 20 - 3 \cr = - 11 \cr \cr \Rightarrow \boxed{f\left( { - 2} \right) = - 11} \cr \cr f\left( {\dfrac{1}{2}} \right) = 3 \cdot {\left( {\dfrac{1}{2}} \right)^2} + 10 \cdot \left( {\dfrac{1}{2}} \right) - 3 \cr = 3 \cdot \dfrac{1}{4} - 5 - 3 \cr = \dfrac{3}{4} - 5 - 3 \cr = - \dfrac{{29}}{4} \cr \cr \Rightarrow \boxed{f\left( {\dfrac{1}{2}} \right) = - \dfrac{{29}}{4}} }\]