Question

4d) write the equation of a rational function with the given characteristics.Vertical asymptote at x = 2, slant asymptote at y =× + 2, passes through the point (3,8)

Answer 1

The rational function will have the next form:

$y=\frac{p(x)}{d(x)}$

where p(x) and d(x) are two polynomials.

Given that the function has a vertical asymptote at x = 2, then d(x) must have a zero at x = 2. Since this is the only asymptote, then:

$d(x)=x-2$

A slant asymptote is present when the degree of the polynomial of the numerator is exactly one more than the degree of the denominator. This means that p(x) must be a polynomial a degree 2.

The quotient, q(x), between p(x) and d(x) is the equation of the slant asymptote. The next relation must be satisfied:

$p(x)=d(x)\cdot q(x)+r(x)$

where r(x) is the remainder of the division. Assuming the remainder is a constant, k, and substituting with d(x) = x-2 and q(x) = x + 2, the slant asymptote, we get:

$\begin{gathered} p(x)=(x-2)(x+2)+k \\ p(x)=(x^2-2^2)+k \\ p(x)=x^2-4+k \end{gathered}$

We know that the curve passes through the point (3, 8), that is, when x = 3, then y = 8. Substituting with this point and the functions p(x) and d(x), we get:

$\begin{gathered} y=\frac{x^2-4+k}{x-2} \\ 8=\frac{3^2-4+k}{3-2} \\ 8=\frac{5+k}{1} \\ 8-5=k \\ 3=k \end{gathered}$

Finally, the rational function is:

$\begin{gathered} y=\frac{x^2-4+3}{x-2} \\ y=\frac{x^2-1}{x-2} \end{gathered}$