Question

Exercitiile 29,30,31 va rog

Answer 1

$\it 29)\ \ \ f(x)=x^2+ax+b\\ \\ V(1,-4) \Rightarrow \begin{cases}\it -\dfrac{a}{2}=1 \Rightarrow a=-2\ \ \ \ \ (*)\\ \\ \\ \it -\dfrac{\Delta}{4}=-4 \Rightarrow \Delta=16 \Rightarrow a^2-4b=16 \stackrel{(*)}{\Longrightarrow}4-4b=16 \Rightarrow b=-3\end{cases}\\ \\ \\ f(x)=x^2-2x-3 \Rightarrow f(2)=2^2-2\cdot2-3=4-4-3=-3$

$\it 30)\ \ f(x)=x^2+(a-2)x+b+3\\ \\ f(0)=6 \Rightarrow b+3=6 \Rightarrow b=3\\ \\ Dreapta\ x=1\ e\ ax\breve a\ de\ simetrie\ pentru\ Gf \Rightarrow \\ \\ \Rightarrow dreapta\ x=1\ trece\ prin\ v\hat arful\ parabolei \Rightarrow -\dfrac{a-2}{2}=1 \Rightarrow a=0\\ \\ f(x)=x^2-2x+6$