that's because the range is all real numbers
There's a vertical asymptote at x=1 which goes to -∞
After that, for x>1, the curve approaches y=x^2+x, which rises to +∞
As x→1, y→ -3, but near x=1, the asymptote takes over, and heads of down to -∞
Take a look at the graph at wolframalpha.com
how do u find the range for this equation
y=x^2+x-2/|x-1|
I went on symbolab, it says range (-3,∞), but I dont understand it
thanks
5 answers
I agree with oobleck
Did you perhaps mean:
y = (x^2 + x - 2)/|(x-1| ??
in that case
y = (x+2)(x-1)/|x-1|
y = x+2 or y = -x-2, but that also would not give you the answer of (-3,∞)
Did you perhaps mean:
y = (x^2 + x - 2)/|(x-1| ??
in that case
y = (x+2)(x-1)/|x-1|
y = x+2 or y = -x-2, but that also would not give you the answer of (-3,∞)
Actually, in this case,
for all x ≠ -1
y = (x+2)(x-1)/|x-1| =
x+2 for x > 1 ... range is (3,∞)
-x-2 for x < -1 ... range is (-3,∞)
So the range here is (-3,∞)
for all x ≠ -1
y = (x+2)(x-1)/|x-1| =
x+2 for x > 1 ... range is (3,∞)
-x-2 for x < -1 ... range is (-3,∞)
So the range here is (-3,∞)
hi, also I understand the line x-int and y-int , but what about the other line? how did u find that
of course!
For some weird reason I was looking at the domain.
For some weird reason I was looking at the domain.