Question

69-74 Use the guidelines of this section to sketch the curve, In guideline D, find an equation of the slant asymptote.

70. \(y = \frac{{{\bf{1}} + {\bf{5}}x - {\bf{2}}{x^2}}}{{x - {\bf{2}}}}\)

Short answer

The sketch of the curve is shown below:

The equation of slant asymptote is \(y = - 2x + 1\).

Step-by-step solution

Find domain using guideline A

The equation\(y = \frac{{1 + 5x - 2{x^2}}}{{x - 2}}\)can be written as:

\(\begin{array}{c}y = \frac{{1 + 5x - 2{x^2}}}{{x - 2}}\\ = - 2x + 1 + \frac{3}{{x - 2}}\end{array}\)

The above equation will be defined, if;

\(\begin{array}{c}x - 2 \ne 0\\x \ne 2\end{array}\)

So, the domain of the function is \(\left( {\infty ,2} \right) \cup \left( {2,\infty } \right)\).

Find intercepts using guideline B

For\(f\left( x \right) = 0\), by the equation of curve\(y = 0\)(x-intercept).

\(\begin{array}{c}1 + 5x - 2{x^2} = 0\\x = \frac{{ - 5 \pm \sqrt {33} }}{{ - 4}}\\ \approx - 0.19,\;{\rm{2}}{\rm{.69}}\end{array}\)

For\(x = 0\), by the equation of curve:

\(f\left( 0 \right) = - \frac{1}{2}\)

Check symmetry using guideline C

Find the value of\(f\left( { - x} \right)\).

\(\begin{array}{c}f\left( { - x} \right) = \frac{{1 + 5\left( { - x} \right) - 2{{\left( { - x} \right)}^2}}}{{\left( { - x} \right) - 2}}\\ = - \frac{{1 - 5x - 2{x^2}}}{{x + 2}}\end{array}\)

So, there is no symmetry.

Find asymptotes guideline D

The vertical asymptote to the curve of\(y = f\left( x \right)\).

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = - \infty \)and\(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \infty \)

So,\(x = 2\)is a vertical asymptote.

Find slant asymptote for the curve of\(y = f\left( x \right)\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \pm \infty } \left( {f\left( x \right) - \left( { - 2x + 1} \right)} \right) = \mathop {\lim }\limits_{x \to \pm \infty } \frac{3}{{x - 2}}\\ = 0\end{array}\)

So, the line \(y = - 2x + 1\) is a slant asymptote.

Find intervals of decrease and increase using guideline E

Differentiate the equation\(f\left( x \right) = - 2x + 1 + \frac{3}{{x - 2}}\).

\(\begin{array}{c}f'\left( x \right) = - 2 - \frac{3}{{{{\left( {x - 2} \right)}^2}}}\\ = \frac{{ - 2\left( {{x^2} - 4x + 4} \right) - 3}}{{{{\left( {x - 2} \right)}^2}}}\\ = \frac{{ - 2{x^2} + 8x - 11}}{{{{\left( {x - 2} \right)}^2}}}\end{array}\)

\(f'\left( x \right) < 0\), for\(x \ne 2\), so f is decreasing on \(\left( { - \infty ,2} \right)\) and \(\left( {2,\infty } \right)\).

Find local maximum or minimum values using guideline F

The root of the equation\(f'\left( x \right) = 0\)does not exist.

So, there is no local extrema.

Find concavity and point of inflection using guideline G

Differentiate the equation\(f'\left( x \right) = \frac{{ - 2{x^2} + 8x - 11}}{{{{\left( {x - 2} \right)}^2}}}\).

\(\begin{array}{c}f''\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{ - 2{x^2} + 8x - 11}}{{{{\left( {x - 2} \right)}^2}}}} \right)\\ = \frac{{{{\left( {x - 2} \right)}^2}\left( { - 4x + 8} \right) - \left( { - 2{x^2} + 8x - 11} \right)\left( {2\left( {x - 2} \right)} \right)}}{{{{\left( {x - 2} \right)}^4}}}\\ = \frac{6}{{{{\left( {x - 2} \right)}^3}}}\end{array}\)

\(f''\left( x \right) > 0\), for\(x > 2\)and\(f''\left( x \right) < 0\), for\(x < 2\).

Therefore, curve has concave upward on \(\left( {2,\infty } \right)\) and concave downward on \(\left( { - \infty ,2} \right)\).

Sketch the curve of the equation \(y = \frac{{{\bf{1}} + {\bf{5}}x - {\bf{2}}{x^2}}}{{x - {\bf{2}}}}\)

The figure below represents the curve of the function \(y = \frac{{1 + 5x - 2{x^2}}}{{x - 2}}\).