What is the area of the polygon
1 answer:
You might be interested in
Zeroes
set numerator to zero
x^2=0
x=0
zeroes are at (0,0)
set numerator to zero
x^2=0
x=0
zeroes are at (0,0)
HA
for p(x)/q(x)
when degree of p(x)<q(x), HA=0
when degree of p(x)=q(x), HA= leading coef of p(x) divided by leading coef of q(x)
when degree of p(x)>q(x) you probably have a slant assymtote
degrees are same (no slant assymtote)
leading coefs
1/1=1
HA is y=1
does it cross?
1=(x^2)/(x^2-4)
x^2-4=x^2
minus x^2 both sides
-4=0
false, it does not cross the HA
VA
simplify the fraction (it can't)
set the denomenator to zero
(x-2)(x+2)=0
VA's at x=2 and x=-2
so to graph
graph the point (0,0)
draw the lines
y=1
x=-2
x=2
reemmber, plus, minus, plus
so from left, it goes from above the HA right up to the VA of x=-2
then goes upside down U shape in between VA's going through (0,0)
then from top of VA x=2 down to y=1
then gets closer to y=1 but never touches