Question

47-52: Find the horizontal and vertical asymptotes of each curve. You may want to use a graphing calculator (or computer) to check your work by graphing the curve and estimating the asymptotes.

49. \(y = \frac{{2{x^2} + x - 1}}{{{x^2} + x - 2}}\)

Short answer

The line \(y = 2\) is a horizontal asymptote of the curve. The vertical lines \(x = - 2\) and \(x = 1\) is the vertical asymptotes of the curve.

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 horizontal and vertical asymptotes of the curve

Divide both numerator and denominator by the highest power of\(x\)in the denominator and evaluate the horizontal asymptotes as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \pm \infty } \frac{{2{x^2} + x - 1}}{{{x^2} + x - 2}} = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{\frac{{2{x^2} + x - 1}}{{{x^2}}}}}{{\frac{{{x^2} + x - 2}}{{{x^2}}}}}\\ &= \mathop {\lim }\limits_{x \to \pm \infty } \frac{{2 + \frac{1}{x} - \frac{1}{{{x^2}}}}}{{1 + \frac{1}{x} - \frac{2}{{{x^2}}}}}\\ &= \frac{{\mathop {\lim }\limits_{x \to \pm \infty } \left( {2 + \frac{1}{x} - \frac{1}{{{x^2}}}} \right)}}{{\mathop {\lim }\limits_{x \to \pm \infty } \left( {1 + \frac{1}{x} - \frac{2}{{{x^2}}}} \right)}}\\ &= \frac{{\mathop {\lim }\limits_{x \to \pm \infty } 2 + \mathop {\lim }\limits_{x \to \pm \infty } \frac{1}{x} - \mathop {\lim }\limits_{x \to \pm \infty } \frac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \pm \infty } 1 + \mathop {\lim }\limits_{x \to \pm \infty } \frac{1}{x} - 2\mathop {\lim }\limits_{x \to \pm \infty } \frac{1}{{{x^2}}}}}\\ &= \frac{{2 + 0 - 0}}{{1 + 0 - 2\left( 0 \right)}}\\ &= 2\end{aligned}\)

Therefore, the line\(y = 2\)is a horizontal asymptote of the curve.

Write the equation as shown below:

\(\begin{aligned}y &= f\left( x \right)\\ &= \frac{{2{x^2} + x - 1}}{{{x^2} + x - 2}}\\ &= \frac{{\left( {2x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}\end{aligned}\)

Determine the vertical asymptotes of the curve as shown below:

\

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - {2^ - }} f\left( x \right) = \infty \left( {{\mathop{\rm since}\nolimits} \,\left( {2x - 1} \right)\left( {x + 1} \right) \to 5\,\,\,{\mathop{\rm and}\nolimits} \,\,\left( {x + 2} \right)\left( {x - 1} \right) \to {0^ - }\,\,{\mathop{\rm as}\nolimits} \,\,x \to - {2^ - }} \right)\\\mathop {\lim }\limits_{x \to - {2^ + }} f\left( x \right) = - \infty \left( {{\mathop{\rm since}\nolimits} \,\left( {2x - 1} \right)\left( {x + 1} \right) \to 5\,\,\,{\mathop{\rm and}\nolimits} \,\,\left( {x + 2} \right)\left( {x - 1} \right) \to {0^ + }\,\,{\mathop{\rm as}\nolimits} \,\,x \to - {2^ + }} \right)\\\mathop {\lim }\limits_{x \to {1^{^ - }}} f\left( x \right) = - \infty \left( {{\mathop{\rm since}\nolimits} \,\left( {2x - 1} \right)\left( {x + 1} \right) \to 2\,\,{\mathop{\rm and}\nolimits} \,\,\left( {x + 2} \right)\left( {x - 1} \right) \to {0^ - }\,\,{\mathop{\rm as}\nolimits} \,\,x \to {1^ - }} \right)\\\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \infty \left( {{\mathop{\rm since}\nolimits} \,\left( {2x - 1} \right)\left( {x + 1} \right) \to 2\,\,{\mathop{\rm and}\nolimits} \,\,\left( {x + 2} \right)\left( {x - 1} \right) \to {0^ + }\,\,{\mathop{\rm as}\nolimits} \,\,x \to {1^ + }} \right)\end{array}\)

Therefore, the vertical line \(x = - 2\) and \(x = 1\) is the vertical asymptotes of the curve.

Check the answer by graphing the curve

The procedure to draw the graph of the equation by using the graphing calculator is as follows:

To check the answer, draw the graph of the function\(f\left( x \right) = \frac{{2{x^2} + x - 1}}{{{x^2} + x - 2}}\)by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation \(\left( {2{X^2} + X - 1} \right)/\left( {{X^2} + X - 2} \right)\) in the \({Y_1}\) tab.
2. Enter the equation\(y = 2\)in the second tab.
3. Enter the equation\(x = - 2\)in the third tab.
4. Enter the equation\(x = 1\)in the fourth tab.
5. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function \(f\left( x \right) = \frac{{2{x^2} + x - 1}}{{{x^2} + x - 2}}\) as shown below:

It is observed from the graph that the line \(y = 2\) is the horizontal asymptote and \(x = - 2\)\(x = 1\) are the vertical asymptotes.