Two parabola sketches: algebra and calculator branches

$y^2=6x+12=6[x-(-2)]$. Comparing $(y-k)^2=a(x-h)$ gives vertex $(-2,0)$ and symmetry $y=0$. The parabola opens right. At $x=0$, $y=\pm\sqrt{12}=\pm2\sqrt3$; at $y=0$, $x=-2$. Both upper and lower branches belong to the curve.

$y=-\dfrac{x^2}{4}-1=-\dfrac14(x-0)^2+(-1)$. The vertex is $(0,-1)$, the graph opens down and its line of symmetry is $x=0$. The only axial intercept is $(0,-1)$: $y=0$ would require $x^2=-4$, so there is no real $x$-intercept.

For the source’s function-graphing workflow, first make $y$ the subject: $y=\pm\sqrt{12+6x}$. In the source table, press y= and enter $Y_1=\sqrt{12+6x}$ and $Y_2=-\sqrt{12+6x}$, then graph. Both have real domain $x\ge-2$. Use the displayed 2nd, trace workflow to locate intercepts numerically. The source explicitly warns that those readings are non-exact: retain $(0,\pm\sqrt{12})$ as exact intercepts. Its two screen displays show these two entries, then their joined rightward parabola; the independently drawn solution figure preserves that mathematics rather than reproducing the calculator interface.