Lets say we have
P(x)/q(x)
vertical assymtotes are in the form x=something, not y=0
y=0 are horizontal assemtotes
so verticall assymtotes
reduce the fraction
set the denomenator equal to zero
those values that make the deomenator zero are the vertical assymtotes
the horizontal assymtote
when the degree of P(x)<q(x), then HA=0
when the degree of P(x)=q(x), then divide the leading coefient of P(x) by the leading coeficnet of q(x)
example, f(x)=(2x^2-3x+3)/(9x^2-93x+993), then HA is 2/9
ok so for vertical assymtote example
f(x)=x/(x^2+5x+6)
the VA's are at x=-3 and x=-2
horizontal assymtote
make degree same
f(x)=(3x^2-4)/(8x^2+9x),
the HA is 3/8
hope I helped, read the whole thing then ask eusiton
Answer:
C. 
Step-by-step explanation:
You are given the exponential function 
From the table, 
at 
thus
![N(0)=a\cdot b^0\\ \\150=a\cdot 1\ [\text{ because }b^0=1]](https://tex.z-dn.net/?f=N%280%29%3Da%5Ccdot%20b%5E0%5C%5C%20%5C%5C150%3Da%5Ccdot%201%5C%20%5B%5Ctext%7B%20because%20%7Db%5E0%3D1%5D)
Also 
at 
thus
Since 
substitute it into the second equation
and the expression for the exponential function is
Answer:
The constants are 7 and 1
Step-by-step explanation:
