y=1-x² y=x²-1 plis yg tau jwb 1 lgi y=-√x (x-3)²+y²=4

Atas.
[tex]1-x^2=x^2-1 \\ 2x^2=2 \\ x^2=1 \\ x^2-1=0 \\ (x+1)(x-1)=0 \\ x_1=-1 \\ y_1=1-x_1^2=1-(-1)^2=1-1=0 \\ x_2=1 \\ y_2=1-x_2^2=1-(1)^2=1-1=0 \\ \\ HP=\{(x,y)|(-1,0),(1,0)\}[/tex]

Bawah:
[tex]y=-\sqrt{x} \\ (x-3)^2+y^2=4 \\ (x-3)^2+(-\sqrt{x})^2=4 \\ x^2-6x+9+x^2=4 \\ 2x^2-6x+5=0 \\ D<0 \\ $Tidak ada penyelesaian, maka:$ \\ HP=\{\}[/tex]