Функция задана формулой f(x)=1/3x^2+8x a)f(3),f(-1) б)нули функции

[latex]f(x)=\frac{1}{3}x^2+8x\\ \\ f(3)=\frac{1}{3}*3^2+8*3=27\\ f(-1)=\frac{1}{3}(-1)^2+8*(-1)=-7\frac{2}{3}\\ \\ \frac{1}{3}x^2+8x=0\\ x(\frac{1}{3}x+8)=0\\ \left \{ {{x=0} \atop {\frac{1}{3}x+8=0}} \right. \\ \left \{ {{x=0} \atop {\frac{1}{3}x=-8}} \right. \\ \left \{ {{x=0} \atop {x=-24}} \right. \\[/latex]