Question

The ration function $f\left(x\right)$ with a hole at $x=5$, a vertical asymptotes at $x=-1$, a horizontal asymptote at $y=2$ and an $x$ intercept at $x=2$

• A. $f\left(x\right)=\frac{2(x-5)(x-2)}{(x+1)(x-5)}$
• B. $f\left(x\right)=\frac{2(x-8)(x-2)}{(x+1)(x-5)}$
• C. $f\left(x\right)=\frac{2(x-5)(x-2)}{(x+8)(x-9)}$
• D. $f\left(x\right)=\frac{(x-5)(x-2)}{(x+1)(x-5)}$

Answer 1

The correct option is A $f\left(x\right)=\frac{2(x-5)(x-2)}{(x+1)(x-5)}$

Let the function $f\left(x\right)=\frac{a{x}^2+bx+c}{a^{\prime }{x}^2+b^{\prime }x+c^{\prime }}$
Since the degree of numerator is equal to the degree of denominator, hence the horizontal asymptote is given by $y=\frac{a}{a^{\prime }}=\mathrm{2...}\left(given\right)$. Thus $a=2a^{\prime }$.
Hence
$f\left(x\right)=\frac{2a^{\prime }{x}^2+bx+c}{a^{\prime }{x}^2+b^{\prime }x+c^{\prime }}$. Now $x_{intercept}$ is given by $y=0$. Thus at $x=2$, $y=0$. Also, at $x=5$ $y=0$. Hence the numerator can be written as
$f\left(x\right)=\frac{(x-2)(x-5)}{a^{\prime }{x}^2+b^{\prime }x+c}=\frac{(x^2-7x+10)}{a^{\prime }{x}^2+b^{\prime }x+c}$. Hence $2a^{\prime }=1$ or $a^{\prime }=\frac{1}{2}$
Hence
$f\left(x\right)=\frac{(x^2-7x+10)}{\frac{1}{2}{x}^2+b^{\prime }x+c}$
$=\frac{2(x^2-7x+10)}{x^2+2b^{\prime }x+2c}$.
$=\frac{2(x^2-7x+10)}{x^2+Bx+C}$.
Also it is given that $x=-1$ is a vertical asymptote. We get a vertical asymptote at the point where the denominator is equal to 0. Hence $(x+1)$ is a factor of $x^2+Bx+C$. Thus $x^2+Bx+C=(x+1)(x+\lambda )$ where $\lambda$ is a real constant.
Hence the function can be re-written as
$f\left(x\right)=\frac{2(x^2-7x+10)}{(x+1)(x+\lambda )}=\frac{2(x-5)(x-2)}{(x+1)(x+\lambda )}$. Now since there is a hole at $x=5$, hence $(x-5)$ is a factor common to both numerator and denominator. Thus $\lambda =-5$.
Hence
$f\left(x\right)=\frac{2(x-5)(x-2)}{(x+1)(x-5)}$