Let that be

Two vertical asymptotes at -1 and 0

If we simply
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- Denominator has degree 2
- Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
- But there should be a x intercept at -3 so one zeros should be -3
Find a equation
Find zeros
Horizontal asymptote
So our equation is

Graph attached
2r + 2s = 50
2r - s = 17
-----------subtract
3s = 33
s = 11
2r - s = 17
2r - 11 = 17
2r = 28
r = 14
so r = 14 and s = 11
answer is C. r = 14, s = 11
3(g-3)=6
3g-9=6
3g=15
g=5
that's the answer
Length=3width-5
(2(x))+(2(3x-5)=46
2x+6x+10=46
8x=56
x=7
width=7 length=16