Respuesta :

dalendrk
[tex]f(x)=ax^2+bx+c\\\\(0;\ 3)\to x=0;\ y=3\to f(0)=3\\therefore\\3=a\cdot0^2+b\cdot0+c\\\boxed{c=3}\\\\(1;\ 4)\to x=1;\ y=4\to f(1)=4\\therefore\\4=1^2\cdot a+1\cdot b+3\\4=a+b+3\ \ \ |subtract\ 3\ from\ both\ sides\\(1)\boxed{a+b=1}\\\\(-1;-6)\to x=-1;\ y=-6\to f(-1)=-6\\therefore\\-6=(-1)^2a+(-1)b+3\\-6=a-b+3\ \ \ \ |subtract\ 3\ from\ both\ sides\\(2)\boxed{a-b=-9}[/tex]

We have (1) & (2):

[tex]\underline{+\left\{\begin{array}{ccc}a+b=1\\a-b=-9\end{array}\right}\ \ \ \ |add\ both\ sides\\.\ \ \ \ \ \ 2a=-8\ \ \ \ |divide\ both\ sides\ by\ 2\\.\ \ \ \ \ \ \ \boxed{a=-4}\\\\subtitute\ the\ value\ of\ "a"\ to\ the\ first\ equation\\\\-4+b=1\ \ \ \ |add\ 4\ to\ both\ sides\\\boxed{b=5}\\\\Answer{\boxed{\boxed{f(x)=-4x^2+5x+3}}[/tex]

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