Obtain the standard form

$z^2-4 x^2-2y^2-4 z\+8 x-4 y-6\=0$$\stackrel{\text{manipulate equation}}{\to }$$\frac{1}{4}{\left(z-2\right)}^2-\frac{1}{2}{\left(y\+1\right)}^2-{\left(x-1\right)}^2\=1$

Obtain the equivalent of the surfaces in Figures 3.3.8(a, b)

$z^2-4 x^2-2y^2-4 z\+8 x-4 y-6\=0$$\to$

Define $f$ so that the graph of $f\=0$ is a quadric surface

$f≔z^2-4 x^2-2y^2-4 z\+8 x-4 y-6\:$

Complete the square and put $f$ into standard form

$\mathrm{Student}:-\mathrm{Precalculus}:-\mathrm{CompleteSquare}\left(f\=0\,\left[x\,y\,z\right]\right)$

$-\left(-4\right)$

$\/4$

Obtain the equivalent of the surfaces drawn in Figures 3.3.8(a, b)

$\mathrm{plots}:-\mathrm{implicitplot3d}\left(f\=0\,x\=-3..5\,y\=-5..3\,z\=-4..8\,\mathrm{style}\=\mathrm{surfacecontour}\,\mathrm{scaling}\=\mathrm{constrained}\, \mathrm{axes}\=\mathrm{frame}\,\mathrm{tickmarks}\=\left[4\,4\,6\right]\,\mathrm{orientation}\=\left[-50\,60\,0\right]\,\mathrm{grid}\=\left[15\,15\,15\right]\,\mathrm{lightmodel}\=\mathrm{none}\right)$