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Sketching Rational curves

for the equation y=(2x^2 - 12)/(4x^2- 5x -6) how am I supposed to know the right side of the graph approaches the asymptote from the top?

Reply 1

arch17

Universities Forum Helper

Original post by Snoyw

for the equation y=(2x^2 - 12)/(4x^2- 5x -6) how am I supposed to know the right side of the graph approaches the asymptote from the top?

use desmos.com it has a graphing calculator

Reply 2

mqb2766

Original post by Snoyw

for the equation y=(2x^2 - 12)/(4x^2- 5x -6) how am I supposed to know the right side of the graph approaches the asymptote from the top?

Id probably sketch both quadratics so
2x^2-12
and
4x^2-5x-6
The denominator has two roots at 2 and -3/4 (vertical asymptotes of y) and the numerator has roots +/-sqrt(6) which correspond to the x-axis crossing points for y. For large |x| y is positive (horizontal asymptotes of y=1/2 as its 2x^2/4x^2=2/4=1/2).

So starting at the right (large x) say, there is a horzional asymptote at y=1/2 and it crosses the x axis at x=sqrt(6) and goes negative, vertical asymptote at x=2. Similar arguments for starting from the left. Then that just leaves between -3/4 and 2 where both quadratics are negative so y is positive "u" shape with positive vertical asymptotes at -3/4 and 2 and crosses the y axis at 2 (-12/-6).

Note once you had the factors of the denominator, you could have used partial fractions which would make the above arguments a bit clearer.

(edited 11 months ago)

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