How do you find the vertical, horizontal or slant asymptotes for #f(x) = (3x^4 + 2x^2 + 1) /( 5x^4 + x -1)#?

1 Answer
Jan 21, 2017

Answer:

Horizontal : #larr y = 0.6 rarr#
Vertical : #uarr x= -0.7715 darr and uarr x = 0.5469 darr#. The Socratic graphs reveal zeros of #5x^4+x-1#.

Explanation:

y= quotient from division = 3/5 gives an asymptote. The other

asymptotes are given by x = a zero of the denominator,

The graphs are on ad hoc ( for the purpose ) scales.

The last graphs is for the given function and the horizontal

asymptote y = 0.6.

The first three approximate zeros of #5x^4+x-1#.

Note near-linearity in the narrowed x-rang, about the the located

zero of the denominator.

graph{5x^4+x-1 [-1.25, 1.25, -0.625, 0.625]}

graph{5x^4+x-1 [-.772, -.771, -.1, .1]}

graph{5x^4+x-1 [.5468, .547, -0.625, 0.625]}

graph{((3x^4+2x^2+1)/(5x^4+x-1)-y)(y-.6)=0 [-1, 1, -29, 29]}