Describing end behavior
f(x) = (3-x)/(x^2-16)
Describe the behavior of f(x) to the left and right of each asymptote.
Asymptotes are y= 4 and y= -4
The right behavior of the first asymptote at y = 4 is ∞correct? Would the left behavior at asymptote then be - ∞?
And the right behavior of the asymptote y = -4 would also be ∞ and the left would again be -∞?
Would someone mind correcting this if necessary?
http://www.walterzorn.com/grapher/grapher_e.htm -here is a grapher if you need.
Thanks so much!
murphie, this graph may help you visualize the graph .. . .
I'm confused, the graph from my calculator looks completely different, your graph suggests that the left behavior at y=-4 is - infinity, and - infinity on the right?? and for the asymptote at y=4 the left is - infinity and right is infinity?
I am fairly bad at calculus so feel free to correct me if i said something stupid. But thanks for your help.
your equation can be re-arranged into this,
y = (3 - x)/((x - 4)(x + 4)), that means x = -4 and x = 4 are vertical asymptotes.
y approaches zero as x approaches positive or negative infinity