Question

1. Для функцiї f(x) = x² + x знайдiть:
1) f(0);
2) f(-1);
3) f(2);
4) f(-3).​

Answer 1

$f(x)=x^2+x\\\\f(0) =0^2 + 0 =0 + 0 = 0 \\\\ f( - 1) =\left( -1 \right)^2 + \left( -1 \right) = 1 - 1 = 0 \\\\ f(2) =2^2 + 2 =4 + 2 = 6 \\\\ f( - 3) =\left( -3 \right)^2 + \left( -3 \right) = 9 - 3 = 6$