The length of a rectangle is 2 centimeters less than its width. What are the dimensions of the rectangle if it’s area is 120 squ
are centimeters
2 answers:
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First, simplify fraction
(x+4)/((x-2)(x+3))
good
set numerator to zero
x+4=0
x=-4
VA
set denom to zero
(x-2)(x+3)=0
VA's at x=-3 and x=2
HA
for p(x)/q(x)
degreee of p(x)<q(x)
HA=0
no slant asymtote
does it cross HA?
0=(x+4)/(x^2+x-6)
0=x+4
-4=x
it crosses at (-4,0)
to graph
plot point (-4,0)
draw assymtotes
y=0
x=2
x=-3
go from top left to middle botttom to top right
reverse 'L' on top left going through (-4,0)
upside down 'U' at middle
'L' at right top
(x+4)/((x-2)(x+3))
good
set numerator to zero
x+4=0
x=-4
VA
set denom to zero
(x-2)(x+3)=0
VA's at x=-3 and x=2
HA
for p(x)/q(x)
degreee of p(x)<q(x)
HA=0
no slant asymtote
does it cross HA?
0=(x+4)/(x^2+x-6)
0=x+4
-4=x
it crosses at (-4,0)
to graph
plot point (-4,0)
draw assymtotes
y=0
x=2
x=-3
go from top left to middle botttom to top right
reverse 'L' on top left going through (-4,0)
upside down 'U' at middle
'L' at right top