SOLUTION: This is what I have and I did something wrong. Any help would be appreciated. It wouldn't let me put my graph on here, but I have the points displayed. f(x)=-2/(x^2+2x-3) =(

Question 434871: This is what I have and I did something wrong. Any help would be appreciated. It wouldn't let me put my graph on here, but I have the points displayed.
f(x)=-2/(x^2+2x-3)
=(x+3)(x-1)
x=-3 or x=1
Domain is-3< x <1
x y
-3 0
-2 -3
-1 -4
0 -3
1 0

xmin=-4 ymin=-5
xmax=2 ymax=1
xscl=1 yscl=1

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
This is what I have and I did something wrong. Any help would be appreciated. It wouldn't let me put my graph on here, but I have the points displayed.
f(x)=-2/(x^2+2x-3)
=(x+3)(x-1)
x=-3 or x=1
Domain is-3< x <1
x y
-3 0
-2 -3
-1 -4
0 -3
1 0
xmin=-4 ymin=-5
xmax=2 ymax=1
xscl=1 yscl=1
-------------------------------------
Adjusted Answer:
f(x)=-2/(x^2+2x-3)
--
Factor the denominator:
f(x) = -2/[(x+3)(x-1)]
---
The graph will have vertical asymptotes at x = -3 and at x = 1.
---------------------------------------------
Points:
If x = -5, y = -2/[-2*-6] = -1/6
If x = 0, y = -2/[3*-1] = 2/3
If x = 2, y = -2/[5*1] = -2/5
-------------------------------------
Graph:

=======================================================
cheers,
Stan H.
==============

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