Answer :
Given
[tex]-8y=x^2+8x-8[/tex]Solving for y,
[tex]\Rightarrow y=-\frac{1}{8}x^2-x+1[/tex]Therefore, since the coefficient that accompanies the x^2 is negative, the parabola opens downwards.
In general, the vertex and general form of a parabola are
[tex]\begin{gathered} y=a(x-h)^2+k\to\text{vertex form} \\ y=ax^2+bx+c \end{gathered}[/tex]Where (h,k) is the vertex of the parabola
Then,
[tex]\begin{gathered} \Rightarrow ax^2+bx+c=a(x-h)^2+k=ax^2-2ahx+ah^2+k \\ \Rightarrow\begin{cases}a=a \\ b=-2ah \\ c=ah^2+k\end{cases} \end{gathered}[/tex]Finding h and k in our case,
[tex]\begin{gathered} \Rightarrow-1=-2(-\frac{1}{8})h\Rightarrow h=-4 \\ \Rightarrow1=-\frac{1}{8}(-4)^2+k \\ \Rightarrow k=1+2 \\ \Rightarrow k=3 \\ \Rightarrow(h,k)=(-4,3) \end{gathered}[/tex]The vertex of the parabola is (-4,3)
Notice that the parabola opens downwards and it is parallel to the y-axis; therefore,
[tex](x-h)^2=4p(y-k)[/tex]Where (h,k) is the vertex. In our case, from the vertex form of the quadratic equation,
[tex]\begin{gathered} y=-\frac{1}{8}(x+4)^2+3 \\ \Rightarrow(x+4)^2=-8(y-3) \end{gathered}[/tex]Thus,
[tex]\begin{gathered} \Rightarrow4p=-8 \\ \Rightarrow p=-2 \end{gathered}[/tex]The distance from the focus to the vertex is 2.
Because the graph of the function opens downwards, we can find its focus as shown below,
[tex]\begin{gathered} (h,k)=(-4,3) \\ \Rightarrow focus\colon(-4,3-2)=(-4,1)_{} \\ \Rightarrow\text{focus(-4,1)} \end{gathered}[/tex]The focus is (-4,1)
Because the graph opens downward and p=-2, the equation of the directrix is
[tex]\begin{gathered} y=k+|p|=3+2 \\ \Rightarrow y=5 \end{gathered}[/tex]The directrix is y=5
Finally, the axis of symmetry is a line perpendicular to the directrix and crosses the vertex; therefore,
[tex]\begin{gathered} y=5\to\text{ slope=0} \\ (h,k)=(-4,3) \\ \Rightarrow x=-4\to\text{ axis of sy}mmetry \end{gathered}[/tex]The axis of symmetry is x=-4