Rational Function with requirements

Question

How can I construct a function $h(x)$, which meets the following requirements:

• The x-intercepts are at 1 and 3

• The vertical asymptotes are at 2 and 5

• The function has a horizontal asymptote $y=4$?

Answer 1

In order to construct $h(x)$ with $x$-intercepts at $1$ and $3$, you want $h(1)=0$ and $h(3)=0$ to be true. So set the numerator of $h(x)$ equal to $$f(x)=(x-3)(x-1)$$

To give $h(x)$ vertical asymptotes at $2$ and $5$, this would mean $$h(2)=\dfrac{1}{0}$$ and $$h(5)=\dfrac{1}{0}$$ (Note that $\dfrac{1}{0}$ is not actually a real value). So to make this true, we set the denominator equal to $$g(x)=\dfrac{1}{(x-2)(x-5)}$$ To give $h(x)$ a horizontal asymptote at $4$, just multiply $h(x)$ by $4$. Now our function is $$h(x)=\dfrac{4f(x)}{g(x)}=\dfrac{4(x-3)(x-1)}{(x-2)(x-5)}$$ and if you feel the need, you can expand this to get $$h(x)=\dfrac{4(x^{2}-4x+3)}{x^{2}-7x+10}$$

Answer 2

What about $f(x)=\frac{(x-1)(x-3)}{(x-2)(x-5)^2}+4$ ?.