Question 269853
Find an equation of a rational function f that satisfies the conditions:
vertical asymptotes: x = - 2, x = 0
we can write a general form as
(i) {{{y = (ax+b)/(x(x+2)))}}}
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horizontal asymptote: y = 0
This means as x approaches infinity, then the function approaches zero. This happens when the degree of the numerator is less  than the degree of the denominator
we keep our function at
(ii) {{{y = (ax+b)/(x(x+2)))}}}
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x-intercept: 5
If y = 0 , then x = 5.
From (ii) we get
(iii){{{(5a+b)/(5(5+2)) = 0}}}
cross multiply to get
5a+b = 0
solving for a , we get
(iv) {{{a = -b/5}}}
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f (6) = 1
means that if x = 6, then y = 1.
From (i) and (iv) we get
(vi) {{{1 = (6(-b/5)+b)/(6(6+2)))}}}
cross multiplying we get
(vii) {{{48 = -6b/5 + b}}}
multiply by common denominator of 5, we get
(viii) {{{240 = -6b + 5b}}}
solving for b, we get
(ix) {{{b = -240}}} 
This means that
{{{a = 48}}}
and our equation is
{{{y = (48x-240)/(x(x+2)))}}}