Question

homework assignment 5.6 rational functions
score: 4.88/12 answered: 5/12
question 6
write an equation for a rational function with:
vertical asymptotes at ( x = 5 ) and ( x = - 5 )
( x )-intercepts at ( x = 2 ) and ( x = 3 )
horizontal asymptote at ( y = 9 )
( y=)
question help: video read

Explanation:

Step1: Determine the denominator

Since the vertical asymptotes are at \(x = 5\) and \(x=-5\), the denominator of the rational function is \((x - 5)(x + 5)=x^{2}-25\) (using the difference - of - squares formula \((a - b)(a + b)=a^{2}-b^{2}\)).

Step2: Determine the numerator

Since the \(x\) - intercepts are at \(x = 2\) and \(x = 3\), the numerator of the rational function is \(a(x - 2)(x - 3)=a(x^{2}-5x + 6)\) (using the FOIL method \((m - n)(m - p)=m^{2}-(n + p)m+np\)).

Step3: Determine the leading coefficient \(a\)

The horizontal asymptote of a rational function \(y=\frac{f(x)}{g(x)}\) where \(f(x)=a_nx^n+\cdots\) and \(g(x)=b_mx^m+\cdots\) is given by \(y = 0\) if \(n\lt m\), \(y=\frac{a_n}{b_m}\) if \(n = m\), and no horizontal asymptote if \(n\gt m\). Here, the degree of the numerator and the denominator is \(2\). We know that the horizontal asymptote \(y = 9\).
Since the degree of the numerator \(n = 2\) and the degree of the denominator \(m = 2\), and \(y=\frac{a}{1}\) (the leading coefficient of the numerator is \(a\) and the leading coefficient of the denominator is \(1\)). So \(a = 9\).

Answer:

\(y=\frac{9(x - 2)(x - 3)}{(x - 5)(x + 5)}=\frac{9x^{2}-45x + 54}{x^{2}-25}\)