Question

Determinați funcția f:R->R,f(x)=x'2+(3a-1)x-2b

Answer 1

$f(x) = x^2+(3a-1)x-2b \\ \\ Avem V(x_0,y_0) \Leftrightarrow V\Big( -\dfrac{b'}{2a'},- \dfrac{\Delta}{4a'}\Big) \\ \\ \Delta = b'^2-4a'c' = (3a-1)^2-4\cdot1\cdot(-2b) = (3a-1)^2+8b\\ Conditiile sunt: \left\{ \begin{array}{c} x_0= 5 \\ y_0 = 9 \end{array} \right \Rightarrow \left\{ \begin{array}{c} -\dfrac{b'}{2a'}= 5 \\ \\ - \dfrac{\Delta}{4a'} = 9 \end{array} \right \Rightarrow$
$\Rightarrow \left\{ \begin{array}{c} -\dfrac{3a-1}{2\cdot1}= 5 \\ \\ - \dfrac{(3a-1)^2+8b}{4\cdot 1} = 9 \end{array} \right \Rightarrow \left\{ \begin{array}{c} 3a-1= -10 \\ (3a-1)^2+8b}= -36 \end{array} \right \Rightarrow \\ \\ \\ \Rightarrow \left\{ \begin{array}{c} 3a= -9 \\ (-10)^2+8b}= -36 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a = -3 \\ 100+36= -8b \end{array} \right \Rightarrow$

$\Rightarrow \left\{ \begin{array}{c} a = -3 \\ b = - \dfrac{136}{8} \end{array} \right \Rightarrow \left\{ \begin{array}{c} \boxed{a = -3} \\ \boxed{b = -17} \end{array} \right |$

$\Rightarrow \boxed{\boxed{f(x) = x^2+10x+34}}$