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

Oct 18, 2016

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

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