Question

65-68 Find an equation of Slant asymptote. Do not sketch the curve

65. \(y = \frac{{{x^{\bf{2}}} + {\bf{1}}}}{{x + {\bf{1}}}}\)

Short answer

\(y = x - 1\) is the slant asymptote.

Step-by-step solution

Use long division to simplify the equation of y

Divide\({x^2} + 1\)by\(x + 1\)by using long division.

\(x + 1\mathop{\left){\vphantom{1\begin{array}{l}{x^2}\,\,\,\,\,\,\,\,\, + 1\\\underline {{x^2} + x} \\\,\,\,\,\,\,\,\, - x + 1\\\,\,\,\,\,\,\,\,\underline { - x - 1} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\end{array}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^2}\,\,\,\,\,\,\,\,\, + 1\\\underline {{x^2} + x} \\\,\,\,\,\,\,\,\, - x + 1\\\,\,\,\,\,\,\,\,\underline { - x - 1} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\end{array}}}}

\limits^{\displaystyle\,\,\, {x - 1}}\)

Re-write the equation of y and simplify

The equation of y can be written as:

\(\begin{array}{c}y = x - 1 + \frac{2}{{x + 1}}\\f\left( x \right) - \left( {x - 1} \right) = \frac{2}{{x + 1}}\\ = \frac{{\frac{2}{x}}}{{1 + \frac{1}{x}}}\end{array}\)

If \(x \to \pm \infty \), then \(f\left( x \right) - \left( {x - 1} \right) = 0\). Therefore, the line of \(y = x - 1\) is a slant asymptote.