怎么求此题垂直渐近线?(全答者有赏)f(x)=(2x^2+5x-42)/(3x^2-8x-3)1.Findallverticalasymptote(s).x=,x=,andx=.2.Findallx-intercepts.Entereachinterceptasapoint.intercept1:,intercept2:.3.Findthey-interceptofthegraph.Enter

Question

怎么求此题垂直渐近线?(全答者有赏)
f(x)=(2x^2+5x-42)/(3x^2-8x-3)
1.Find all vertical asymptote(s).
x= ,x= ,and x= .
2.Find all x-intercepts.Enter each intercept as a point.
intercept 1:,intercept 2:.
3.Find the y- intercept of the graph.Enter the intercept as a point.
4.Find the horizontal asymptote.
y= .

Answer 1

f(x)=(x+6)(2x-7)/[(x-3)(3x+1)]
1.
当x-->-1/3或者3时,f(x)-->∞.
所以垂直渐近线有两条：x=-1/3,x=3.
2.
与x轴交点有两个：
(-6,0),(7/2,0）.
3.
与y轴交点是：
(0,14)
4.
lim{x-->∞}f(x)
=lim{x-->∞}(2+5/x-42/x^2)/(3-8/x-3/x^2)
=2/3
因此水平渐近线是：y=2/3