[tex]\it P(a,\ \ -a)\in Gf \Rightarrow f(a)=-a \Rightarrow -4a-11=-a \Rightarrow -4a+a=11 \Rightarrow\\ \\ \Rightarrow -3a = 11|_{\cdot(-1)} \Rightarrow 3a=-11 \Rightarrow a=-\dfrac{11}{3}\\ \\ Prin\ urmare\ punctul\ cerut\ este\ P\Big(-\dfrac{11}{3},\ \dfrac{11}{3}\Big)[/tex]
