Question 189950
We have the given function as:

{{{ f(x)=(3x^2-3)/(x^2-16) }}}

>>> (a) Find the domain.

>>> (- infinity, -4) U (-4,4) U (4, infinity) ----- This is perfect and correct.


>>> (b) Determine the vertical asymptote(s).

>>> x= -4 and x= 4, those are the vertical asymptotes! --- Correct

>>> (c) Determine the horizontal asymptote or oblique asymptote. 

>>>> Y = 3/1 -------- Correct

The numerator is dominated by 3x2 and the denominator is dominated by x2

Since it states P(x)/Q(x), it must be 3/1


>>>> (d)  Find the y-intercept.

To find the y-intercept we plug in x=0 in the equation. The concept of x and y-intercepts is that, when the graph/curve/function crosses the x-axis, it cuts t x-axis at some point (a,0), this point is known as the x-intercept, here y-coordinate=0. Similar is the case for y-intercepts. when the curve cuts the y-axis at (0,b) the x-coordinate =0 and this point is known as the y-intercept.

so, to find the y-intercept, we put x=0, we have: (here y = f(x))

{{{ f(x) = (3(0)^2-3)/((0)^2-16) }}}
{{{ f(x) = -3/(-16) }}}
or, {{{ f(x) = 3/16 }}}
so we have the y-intercept as (0,3/16)

(e) Find the x-intercept(s).

so, to find the x-intercept, we put y=0, we have: (here y = f(x))

{{{ 0 = (3x^2-3)/(x^2-16) }}}
{{{ 0 = 3x^2-3 }}}
or, {{{ 3x^2 = 3 }}}  (divide this equation by 3)
=> {{{ x^2 = 1 }}}
=>  x = +1 or x = -1 (after taking the square root)
so we have the x-intercepts as (-1,0) and (1,0)

>>>I tried a-c and hope it is correct, but I don't know how to find the >>>intercepts with this number or I am just confused since we have done >>>intercepts at the beginning of the semester...

Good Luck