Question

How do you find all the critical points to graph `4x^2 + 20x - 4y^2 + 6y - 3 = 0` including vertices, foci and asymptotes?

Answer 1

The graph is a hyperbola and we have to write the equation in the standard form.
`4x^2+20x-4y^2+6y-3=4(x^2+5x+25/4)-4(y^2-3/2y+9/16)-3-25-9/4=0`
So the equation becomes
`4(x+5/2)^2-4(y-3/4)^2=121/4`
`(x+5/2)^2/(121/16)-(y-3/4)^2/(121/16)=1`
So the center is `(-5/2,3/4)`
The vertices are `(-5/2+11/4,3/4)` and `(-5/2-11/4,3/4)` which is `(1/4,3/4)and (-21/4,3/4)`
The slope of the asymptotes are 1 and -1
The equations of the asymptotes are `y=3/4+x+5/2` and `y=3/4-x-5/2` that is `y=x+13/4` and `y=-x-7/4`
In order to determine the foci we need `c=+-sqrt(2*(121/6))=+-11/6sqrt2`
So the foci are `(-5/2+)` and ##

graph{4x^2+20x-4y^2+6y-3=0 [-20, 20, -10, 10]}