How do you graph #y =(x^2+4x-2)/(x^2-x-7)#?

1 Answer
A. S. Adikesavan
Jul 29, 2018

See graph and details.

Explanation:

By actual division,

#y = 1 + ( 5 ( x + 1 ))/(( x - a )(x - b )), a,b = 1/2( 1 +- sqrt29 )#. So,

#( y - 1 )( x - a )( x - b ) = 5 ( x + 1 )#, with the asymptotic

#y ne 1, x ne a and x ne b )#.

See graph, with asymptotes. y = 1, x = a and x = b.
graph{(y(x^2-x-7)-(x^2+4x-2))(y-1)(x-1/2-1/2sqrt29+0.00001y)(x-1/2+1/2sqrt29+0.00001y)=0}

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