Question

срочнооо!!!!.........​

Answer 1

Ответ:

$f(x)=ax^2+bx+c\\\\f(0)=4\ \ ,\ \ f(0)=a\cdot 0^2+b\cdot 0+c=c\ \ ,\ \ c=4\\\\f(-3)=1\ \ ,\ \ f(-3)=9a-3b+4=1\ \ ,\ \ 9a-3b=-3\ \ ,\ \ 3a=b-1\\\\x_{versh.}=-3\ \ ,\ \ x_{versh.}=-\dfrac{b}{2a}=-3\ \ ,\ \ b=6a\\\\\left\{\begin{array}{l}3a=b-1\\b=6a\end{array}\right\ \ \Rightarrow \ \ \ 3a=6a-1\ \ ,\ \ 3a=1\ \ ,\ \ a=\dfrac{1}{3}\ \ ,\ \ b=6\cdot \dfrac{1}{3}=2\\\\\\f(x)=\dfrac{1}{3} x^2+2x+4\\\\f(3)=\dfrac{1}{3}\cdot 3^2+2\cdot 3+4=3+6+4=3+10=13$