Question

КАК ЭТО СДЕЛАТЬ!!!!!????

Answer 1

$1)\ \ A(0;16)\ ,\ B(3;-2)\ ,\ C(4;0)\ \ \ ,\\\\y=ax^2+bx+c\\\\A(0;16):\ \ y(0)=16\ \ \to \ \ \ y(0)=c\ \ \ \to \ \ \ c=16\\\\\\y=ax^2+bx+16\ \ ,\ \ y(3)=-2\ \ ,\ \ y(4)=0\\\\\left\{\begin{array}{l}9a+3b+16=-2\\16a+4b+16=0\end{array}\right\ \ \left\{\begin{array}{l}9a+3b=-18\\16a+4b=-16\end{array}\right\ \ \left\{\begin{array}{l}3a+b=-6\\4a+b=-4\end{array}\right\ \ominus \\\\\\\left\{\begin{array}{l}b=-6-3a\\-a=-2\end{array}\right\ \ \left\{\begin{array}{l}b=-12\\a=2\end{array}\right$

    $\boxed{y=2x^2-12x+16}$

$2)\ \ \ y=ax^2+bx+c\\\\V(2;4)\ \ ,\ \ A(0;0)\ ,\ B(4;0)\\\\V(2;4):\ \ x_{vedrsh}=2=-\dfrac{b}{2a}\ \ \to \ \ \ b=-4a\\\\A(0;0):\ \ y(0)=0\ \ \to \ \ y(0)=c\ \ ,\ \ c=0\ \ \to \ \ y=ax^2+bx\\\\B(4;0):\ \ y(4)=0\ \ \to \ \ y(4)=16a+4b=0\ \ ,\ \ 4a+b=0\ \ ,\ \ b=-4a\\\\y=ax^2-4ax\ \ \to \ \ y_{versh}=y(2)=4\ \ ,\ \ y(2)=4a-8a=4\ \ ,\\\\-4a=4\ \ ,\ a=-1\ \ ,\ \ b=-4a=4\\\\\boxed{y=-x^2+4x}$

Answer 2

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