How do you find all the critical points to graph #x^2/9 - y^2/16 = 1# including vertices?

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Answer 1 · 2018-07-03T21:02:31

Given that

#x^2/9-y^2/16=1#

#x^2/3^2-y^2/4^2=1#

The above equation shows a horizontal hyperbola with horizontal semi axis #a=3# & transverse semi-axis #b=4#

The center of hyperbola #(0, 0)#

The vertices of given hyperbola are #(\pma, 0)\equiv(\pm3, 0)# & #(0, \pmb)\equiv(0, \pm4)#

Asymptotes of given hyperbola

#y=\pmb/ax#

#y=\pm4/3x#

Specify all the four vertices of hyperbola with center at origin & draw both the asymptotes. Then draw both the horizontal branches of hyperbola