SOLUTION: Hello, I will first write the task and then what I have tried! f(x)=(3x^2-3)/x^2-16) (a) Find the domain. (- infinity, -4) U (-4,4) U (4, infinity) (b) Determine t

Algebra ->  Functions -> SOLUTION: Hello, I will first write the task and then what I have tried! f(x)=(3x^2-3)/x^2-16) (a) Find the domain. (- infinity, -4) U (-4,4) U (4, infinity) (b) Determine t      Log On


   

Question 189950: Hello,
I will first write the task and then what I have tried!
f(x)=(3x^2-3)/x^2-16)
(a) Find the domain.
(- infinity, -4) U (-4,4) U (4, infinity)

(b) Determine the vertical asymptote(s).
x= -4 and x= 4, those are the vertical asymptotes!
(c) Determine the horizontal asymptote or oblique asymptote.
Y = 3/1
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.
(e) Find the x-intercept(s).

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...


Answer by cutepiscean5(19) About Me  (Show Source):
You can put this solution on YOUR website!
We have the given function as:
+f%28x%29=%283x%5E2-3%29%2F%28x%5E2-16%29+
>>> (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%28x%29+=+%283%280%29%5E2-3%29%2F%28%280%29%5E2-16%29+
+f%28x%29+=+-3%2F%28-16%29+
or, +f%28x%29+=+3%2F16+
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+=+%283x%5E2-3%29%2F%28x%5E2-16%29+
+0+=+3x%5E2-3+
or, +3x%5E2+=+3+ (divide this equation by 3)
=> +x%5E2+=+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