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:
gkali 53 *622 kalia qqua 5% habwa
Answer: the answer is yes
Step-by-step explanation:
the inequality is underlined so the circle is going to be opened and since x is greater than -1 the line is drawn towards the left.
Answer:
The answer is in the included image
Answer:
perimeter is 12x-6
Step-by-step explanation:
perimeter is the sum of all sides
3(x+2) + 2x+11 + 7x-23
distribute the 3 in the first term, then combine 'like terms'
3x+6 + 2x+11 + 7x-23
3x+2x+7x + 6+11+(-23)
12x-6
