Answer:
The correct option is A.
Step-by-step explanation:
Domain:
The expression in the denominator is x^2-2x-3
x² - 2x-3 ≠0
-3 = +1 -4
(x²-2x+1)-4 ≠0
(x²-2x+1)=(x-1)²
(x-1)² - (2)² ≠0
∴a²-b² =(a-b)(a+b)
(x-1-2)(x-1+2) ≠0
(x-3)(x+1) ≠0
x≠3 for all x≠ -1
So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong
Asymptote:
x-3/x^2-2x-3
We know that denominator is equal to (x-3)(x+1)
x-3/(x-3)(x+1)
x-3 will be cancelled out by x-3
1/x+1
We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....
Take into account, that in general, a cosine function of amplitude A, period T and vertical translation b, can be written as follow:
In the given case, you have:
A = 4
T = 3π/4
b = -3
By replacing you obtain:
Hence, the answer is:
f(x) = 4cos(8/3 x) - 3
Answer:
75,000,012
Step-by-step explanation:
enjoy, i guess
Do you see -8x?
Take half of -8 and square it.
So, (-8)/2 = -4.
(-4)^2 = 16
Now add 16 to both sides.
Answer:
I had this question, photomath works.
Step-by-step explanation:
