Answer:
I suppose that the equation is:
f(x) = 8/x
(You wrote 8x, this is a linear function, so it has no asymptotes, then I assume that the actual function was 8/x)
The y-intercept is the value such that f(x) = 0
Notice that there is no value of x such that 8/x = 0 (if the numerator is never zero, then the quotient can't be zero)
So we do not have an y-intercept.
The asymptote:
Here we have four:
if x tends to zero from the negative side, then:
8/x tends to negative infinity.
if x tends to zero from the positive side, then:
8/x tends to positive infinity.
These two asymptotes can be written as:
[tex]\lim_{n \to \_-0} \frac{8}{x} = -\infty\\ \lim_{n \to \_+0} \frac{8}{x} = +\infty[/tex]
We also have two when x tends to plus infinity (and 8/x goes to zero) and when x tends to negative infinity, such that the function tends to zero again, these two can be written as:
[tex]\lim_{n \to \infty} \frac{8}{x} = 0\\ \lim_{n \to -\infty} \frac{8}{x} = 0[/tex]
Finally, the range:
The only value that y never reaches (only on limits) is:
y = 0
Then the range is the set of all real numbers except the zero, or:
R : { x ∈ R / {0} }
