Question

58: Find a formula for a function that has vertical asymptotes \(x = 1\) and \(x = 3\), and horizontal asymptotes \(y = 1\).

Short answer

The formula for the function is \(f\left( x \right) = \frac{{{x^2}}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\).

Step-by-step solution

Condition for horizontal asymptotes and vertical asymptotes

Thehorizontal asymptote of the curve\(y = f\left( x \right)\)for the line \(y = L\)is\(\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = L\) or \(\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = L\).

Thevertical asymptote of the curve\(y = f\left( x \right)\)for the line \(x = a\)are shown below:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to a} f\left( x \right) = \infty \,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \infty \,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = \infty \\\mathop {\lim }\limits_{x \to a} f\left( x \right) = \infty \,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = - \infty \,\,\,\,\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = - \infty \end{array}\)

Determine the formula for the function

The denominator of the rational function should have factors \(\left( {x - 1} \right)\) and \(\left( {x - 3} \right)\) because the function contains vertical asymptotes \(x = 1\) and \(x = 3\).

The degree of the numerator should be equal to the degree of the denominator, and the ratio of the leading coefficient should be equal to 1 since the function has horizontal asymptotes \(y = 1\).

The possibility for the formula of the function is \(f\left( x \right) = \frac{{{x^2}}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\).

Thus, the formula for the function is \(f\left( x \right) = \frac{{{x^2}}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\).