Question 698158
A hyperbola that is centered at the origin, and opens left and right has an equation of the form
{{{x^2/a^2-y^2/b^2=1}}}
where {{{x}}} cannot be zero (because it would mean {{{-y^2/b^2=1}}}<--->{{{y^2/b^2=-1}}},
but when {{{y=0}}}, {{{x^2/a^2=1}}}<--->{{{x^2=a^2}}}, so {{{x=a}}} or {{{x=-a}}} Those are the x-coordinates of vertices (-a,0) and (a,0),
and the distance between the vertices is the transverse axis = {{{2a}}}
So the equation you need is
{{{highlight(x^2/3^2-y^2/b^2=1)}}} or {{{highlight(x^2/9-y^2/b^2=1)}}}
You can chose a value for {{{b^2}}} to change how "pointy" the vertices are.
The slopes of the asymptotes will be {{{b/a}}} and {{{-b/a}}}