the equation is supposed to be
|(6x^2+5x-3)/(2x^2-1) -3 (the negative three is outside of the fraction) |
I have done one part of this question. I am, however, unsure as to how to set up the graph. Can someone help me?
1. Use a graph to find a number N such that
|(6x^2+5x-3)| |
|------------- -3 | < 0.2 whenever x>N
|(2x^2-1)| |
y=2.8 and y=3.2
My issue is how to find the x, or where the curve crosses at the horizontal line at 2.8.
2 answers
let y = (6x^2 + 5x - 3)/(2x^2 - 1) - 3
we can simplify
= ( 6x^2 + 5x - 3 - 3(2x^2 - 1) )/(2x^2 - 1)
= (5x)/(2x^2 - 1)
there will be 2 vertical asymptotes,
x = ± 1/√2
graph looks like this, the asymptotes are not drawn
http://www.wolframalpha.com/input/?i=simplify+%286x%5E2+%2B+5x+-+3%29%2F%282x%5E2+-+1%29+-+3
let's set 5x/(2x^2 - 1) = .2 = 1/5
25x = 2x^2 - 1
2x^2 - 25x - 1 = 0
x = (25 ± √633)/4
= appr 12.54 or -.04
( I don't know how you got your answers)
so (6x^2+5x-3)/(2x^2-1) -3 < .2
for all x>12.54 or x < -.04
we can simplify
= ( 6x^2 + 5x - 3 - 3(2x^2 - 1) )/(2x^2 - 1)
= (5x)/(2x^2 - 1)
there will be 2 vertical asymptotes,
x = ± 1/√2
graph looks like this, the asymptotes are not drawn
http://www.wolframalpha.com/input/?i=simplify+%286x%5E2+%2B+5x+-+3%29%2F%282x%5E2+-+1%29+-+3
let's set 5x/(2x^2 - 1) = .2 = 1/5
25x = 2x^2 - 1
2x^2 - 25x - 1 = 0
x = (25 ± √633)/4
= appr 12.54 or -.04
( I don't know how you got your answers)
so (6x^2+5x-3)/(2x^2-1) -3 < .2
for all x>12.54 or x < -.04