Question

Дана функция (х)=-4^2-3x+16 найдите:

Answer 1

$f(x)=-4x^{2}-3x+16\\\\f(3)=-4*3^{2}-3*3+16=-4*9-9+16=-36+7=-29\\\\\boxed{f(3)=-29}\\\\f(-1)=-4*(-1)^{2}-3*(-1)+16=-4*1+3+16=-4+19=15\\\\\boxedf(-1)=15\\\\f(\frac{1}{2})=-4*(\frac{1}{2})^{2}-3*\frac{1}{2}+16=-4*\frac{1}{4} -\frac{3}{2}+16=-1-1,5+16=13,5\\\\\boxed{f(\frac{1}{2})=13,5}$