We can do long division or perform "juggling of terms" in the numerator (which is basically introducing and returning of terms).

I will do "juggling" here since this chat box does not allow for long division.

Rewriting x + a as x + b + a - b,

y = (x + b + a - b) divided by (x + b)
y = (x + b) divided by (x + b) plus (a - b) divided by (x + b)
y = 1 + (a - b)/(x + b)

We have two asymptotes.

The vertical one is x = -b, found by setting x + b = 0. This is because the graph does not exist at that value but the graph seems to go towards infinity or negative infinity as x gets closer to this "forbidden" value.

The other one is y = 1, found by setting x + b = infinity (the sum a + b is nothing compared to infinity, so dividing a + b by infinity is as good as zero).

For the intercept with the y-axis. x = 0, so y = a/b.

For the intercept with the x-axis, y = 0, so x = -a.

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The fact that the y-intercept is positive but the x-intercept is negative shows that a/b > 0 and a > 0.

If you pay attention carefully, as x approaches infinity, the graph gets closer to 1, but from somewhere below 1.

So, in the equation y = 1 + (a - b)/(x + b), the numerator a - b has to be negative. Else, if a - b is positive, then (a - b)/(x + b) will be positive as x becomes large. This leads to 1 + positive being greater than 1 which is not in agreement with what I wrote in the previous paragraph.

So, a - b < 0.
a < b
b > a