After graphing the parabola [tex](x+1)^2=2(y-3)[/tex], she can fill the boxes as follows:
Gigi identifies that the vertex of the parabola is [tex](-1, 3)[/tex]. The parabola opens up, and the focus is [tex]\frac{1}{2}[/tex] unit away from the vertex. The directrix is [tex]1[/tex] unit from the focus. The focus is the point [tex](-1, \frac{7}{2})[/tex]. The directrix of the equation is [tex]y=\frac{5}{2}[/tex].
Since Gigi's parabola has the x-term squared, then the parabola has the general formula
[tex](x-h)^2=4p(y-k)[/tex]
So,
[tex]x-h=x+1 \implies h=-1[/tex]
[tex]y-k=y-3\implies k=3[/tex]
The vertex is
[tex](h,k)=(-1,3)[/tex]
The focus is
[tex](h,k+p)=(-1,3+\frac{1}{2})\\=(-1,\frac{7}{2})[/tex]
The directrix is
[tex]y=k-p=3-\frac{1}{2}=\frac{5}{2}[/tex]
The other values can be computed as follows;
Direction the parabola opens: Since the value of p is positive, the parabola opens up.
Distance of focus from vertex: The focus and vertex both have the same x position, so the distance will be in terms of the y-coordinates only. The distance from the focus to the vertex is
[tex]\frac{7}{2}-3=\frac{1}{2}\text{ unit}[/tex]
Distance of directrix to focus: The directrix is the line [tex]y=2.5[/tex]. The distance of the directrix to the focus will be
[tex]\frac{7}{2}-\frac{5}{2}=1 \text{ unit}[/tex]
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