Question

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

Answer 1

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