Steps for Sketching the Graph of the Function

Example 1. Sketch graph of the function $$${f{{\left({x}\right)}}}=\frac{{1}}{{{x}-{1}}}$$$.

Since $$${f{{\left({x}\right)}}}={g{{\left({x}-{1}\right)}}}$$$ where $$${g{{\left({x}\right)}}}=\frac{{1}}{{x}}$$$ then we study function $$${g{{\left({x}\right)}}}=\frac{{1}}{{x}}$$$ and then shift graph one unit to the right.

Function is odd because $$${g{{\left({x}\right)}}}=-{g{{\left(-{x}\right)}}}$$$, so we consider only interval $$${\left[{0},\infty\right)}$$$ and then just reflect result about the origin. Function is not periodic.

There is no y-intercept because $$${g{{\left({0}\right)}}}$$$ is undefined.

There is no x-intercept because there are no points where $$${g{{\left({x}\right)}}}=\frac{{1}}{{x}}={0}$$$.

Since $$$\lim_{{{x}\to{{0}}^{+}}}\frac{{1}}{{x}}=+\infty$$$ then $$${x}={0}$$$ is vertical asymptote.

Since $$$\lim_{{{x}\to\infty}}\frac{{1}}{{x}}={0}$$$ then $$${y}={0}$$$ is horizontal asymptote.

We have that $$${g{'}}{\left({x}\right)}=-\frac{{1}}{{{x}}^{{2}}}$$$. There are no points where derivative equals 0. Point $$${x}={0}$$$ is where derivative doesn't exist.

Since $$${g{'}}{\left({x}\right)}=-\frac{{1}}{{{x}}^{{2}}}<{0}$$$ for all $$${x}$$$ then function is decreasing on $$${\left({0},+\infty\right)}$$$ and there are no extrema according to First Derivative Test.

Second derivative is $$${g{''}}{\left({x}\right)}=\frac{{2}}{{{x}}^{{3}}}$$$. There are no points where $$${g{''}}{\left({x}\right)}={0}$$$ and second derivative doesn't exist when $$${x}={0}$$$.

Since second derivative is positive for $$${x}>{0}$$$ then function is concave upward on $$${\left({0},\infty\right)}$$$.

Point $$${x}={0}$$$ is not point of inflection because $$${x}={0}$$$ is not in the domain of the function.

So, we actually don't have important points, but when $$${x}$$$ approaches 0 from the right function grows without a bound and we can sketch function, taking into account that functions is decreasing and concave upward on $$${\left({0},+\infty\right)}$$$. For better graph take a couple of control points: $$${\left({1},{1}\right)},{\left({1},\frac{{1}}{{2}}\right)},{\left({4},\frac{{1}}{{4}}\right)}$$$.

Since function is odd, reflect it about origin.

Shift the function 1 unit to the right.

Example 2. Sketch graph of the function $$${f{{\left({x}\right)}}}={{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{1}\right)}}^{{3}}$$$.

There is no simpler function that initial function is obtained from.

Function is neither even nor odd and not periodic.

$$${f{{\left({0}\right)}}}={{\left({0}+{2}\right)}}^{{2}}{{\left({0}-{1}\right)}}^{{3}}=-{4}$$$, so y-intercept is -4.

x-intercepts are points where $$${f{{\left({x}\right)}}}={0}$$$: $$${{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{1}\right)}}^{{3}}={0}$$$ which gives $$${x}=-{2}$$$ and $$${x}={1}$$$.

Since function is continuous everywhere, $$$\lim_{{{x}\to-\infty}}{f{{\left({x}\right)}}}=-\infty$$$ and $$$\lim_{{{x}\to\infty}}{f{{\left({x}\right)}}}=\infty$$$ then there are no asymptotes.

We have that $$${f{'}}{\left({x}\right)}={2}{\left({x}+{2}\right)}{{\left({x}-{1}\right)}}^{{3}}+{3}{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{1}\right)}}^{{2}}={\left({x}+{2}\right)}{{\left({x}-{1}\right)}}^{{2}}{\left({2}{x}-{2}+{3}{x}+{6}\right)}=$$$

$$$={\left({x}+{2}\right)}{{\left({x}-{1}\right)}}^{{2}}{\left({5}{x}+{4}\right)}$$$.

So, derivative equals 0 when $$${x}=-{2},{x}={1},{x}=-\frac{{4}}{{5}}$$$.

According to method of intervals number line is divided by stationary points on four intervals.

First interval is $$${\left(-\infty,-{2}\right)}$$$: taking any point from this interval, for example $$${x}=-{3}$$$ we obtain that $$${f{'}}{\left(-{3}\right)}={\left(-{3}+{2}\right)}{{\left(-{3}-{1}\right)}}^{{2}}{\left({5}\cdot{\left(-{3}\right)}+{4}\right)}>{0}$$$, so $$${f{'}}{\left({x}\right)}>{0}$$$ on this interval.

Second interval is $$${\left(-{2},-\frac{{4}}{{5}}\right)}$$$: taking any point from this interval, for example $$${x}=-{1}$$$ we obtain that $$${f{'}}{\left(-{1}\right)}={\left(-{1}+{2}\right)}{{\left(-{1}-{1}\right)}}^{{2}}{\left({5}\cdot{\left(-{1}\right)}+{4}\right)}<{0}$$$, so $$${f{'}}{\left({x}\right)}<{0}$$$ on this interval.

Third interval is $$${\left(-\frac{{4}}{{5}},{1}\right)}$$$: taking any point from this interval, for example $$${x}={0}$$$ we obtain that $$${f{'}}{\left({0}\right)}={\left({0}+{2}\right)}{{\left({0}-{1}\right)}}^{{2}}{\left({5}\cdot{\left({0}\right)}+{4}\right)}>{0}$$$, so $$${f{'}}{\left({x}\right)}>{0}$$$ on this interval.

Fourth interval is $$${\left({1},\infty\right)}$$$: taking any point from this interval, for example $$${x}={3}$$$ we obtain that $$${f{'}}{\left({3}\right)}={\left({3}+{2}\right)}{{\left({3}-{1}\right)}}^{{2}}{\left({5}\cdot{3}+{4}\right)}>{0}$$$, so $$${f{'}}{\left({x}\right)}>{0}$$$ on this interval.

According to First Derivative Test $$${x}=-{2}$$$ is local maximum, $$${x}=-\frac{{4}}{{5}}$$$ is local minimum and $$${x}={1}$$$ is not an extremum.

Second derivative is $$${f{''}}{\left({x}\right)}={2}{\left({x}-{1}\right)}{\left({10}{{x}}^{{2}}+{16}{x}+{1}\right)}$$$.

Roots of the equation $$${10}{{x}}^{{2}}+{16}{x}+{1}={0}$$$ are $$${x}=\frac{{-{8}+{3}\sqrt{{{6}}}}}{{10}}\approx-{0.065}$$$ and $$${x}=\frac{{-{8}-{3}\sqrt{{{6}}}}}{{10}}\approx-{1.535}$$$.

So, points where second derivative equals 0 are $$${x}={1},{x}=-{0.065},{x}=-{1.535}$$$.

Clearly they are all points of inflections because second derivative changes sign at these points.

$$${f{''}}{\left({x}\right)}>{0}$$$ on $$${\left(-{1.535},-{0.065}\right)}\cup{\left({1},\infty\right)}$$$ and $$${f{''}}{\left({x}\right)}<{0}$$$ on $$${\left(-\infty,-{1.535}\right)}\cup{\left(-{0.0065},{1}\right)}$$$.

So we have following "important" points:

Function is increasing on $$${\left(-\infty,-{2}\right)}\cup{\left(-\frac{{4}}{{5}},\infty\right)}$$$ and decreasing on $$${\left(-{2},-\frac{{4}}{{5}}\right)}$$$.

Function is concave upward on $$${\left(-{1.535},-{0.065}\right)}\cup{\left({1},\infty\right)}$$$ and concave downward on $$${\left(-\infty,-{1.535}\right)}\cup{\left(-{0.0065},{1}\right)}$$$.

Note that points $$${\left(-{0.065},-{4.052}\right)}$$$ and $$${\left({0},-{4}\right)}$$$ are very close, so they are almost indistinguishable on graph.

Example 3. Sketch graph of the function $$${f{{\left({x}\right)}}}=\frac{{{{x}}^{{2}}-{5}{x}+{6}}}{{{{x}}^{{2}}+{1}}}$$$.

There is no simpler function that initial function is obtained from.

Function is neither even nor odd and not periodic.

$$${f{{\left({0}\right)}}}=\frac{{{{0}}^{{2}}-{5}\cdot{0}+{6}}}{{{{0}}^{{2}}+{1}}}={6}$$$, so y-intercept is 6.

x-intercepts are points where $$${f{{\left({x}\right)}}}={0}$$$: $$${\left({{x}}^{{2}}-{5}{x}+{6}\right)}={\left({x}-{2}\right)}{\left({x}-{3}\right)}={0}$$$ which gives $$${x}={2}$$$ and $$${x}={3}$$$.

Since function is continuous everywhere, $$$\lim_{{{x}\to\infty}}\frac{{{{x}}^{{2}}-{5}{x}+{6}}}{{{{x}}^{{2}}+{1}}}={1}$$$ then there is only one asymptote: horizontal asymptote $$${y}={1}$$$.

We have that $$${f{'}}{\left({x}\right)}=\frac{{{\left({{x}}^{{2}}-{5}{x}+{6}\right)}'{\left({{x}}^{{2}}+{1}\right)}-{\left({{x}}^{{2}}-{5}{x}+{6}\right)}{\left({{x}}^{{2}}+{1}\right)}'}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}={5}\frac{{{{x}}^{{2}}-{2}{x}-{1}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}}$$$.

Roots of the equation $$${{x}}^{{2}}-{2}{x}-{1}={0}$$$ are $$${x}={1}+\sqrt{{{2}}}\approx{2.41}$$$ and $$${x}={1}-\sqrt{{{2}}}\approx-{0.41}$$$.

So, derivative equals 0 when $$${x}=-{0.41},{x}={2.41}$$$.

According to method of intervals number line is divided by stationary points on three intervals.

First interval is $$${\left(-\infty,-{0.41}\right)}$$$: $$${f{'}}{\left({x}\right)}>{0}$$$ on this interval.

Second interval is $$${\left(-{0.41},{2.41}\right)}$$$:$$${f{'}}{\left({x}\right)}<{0}$$$ on this interval.

Third interval is $$${\left({2.41},\infty\right)}$$$: $$${f{'}}{\left({x}\right)}>{0}$$$ on this interval.

According to First Derivative Test $$${x}=-{0.41}$$$ is local maximum, $$${x}={2.41}$$$ is local minimum.

Second derivative is $$${f{''}}{\left({x}\right)}=-{10}\frac{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{4}{x}+{1}\right)}}}{{{\left({{x}}^{{2}}+{1}\right)}}^{{3}}}$$$.

Roots of the equation $$${{x}}^{{2}}-{4}{x}+{1}={0}$$$ are $$${x}={2}-\sqrt{{{3}}}\approx{0.27}$$$ and $$${x}={2}+\sqrt{{{3}}}\approx{3.73}$$$.

So, points where second derivative equals 0 are $$${x}=-{1},{x}={0.27},{x}={3.73}$$$.

Clearly they are all points of inflections because second derivative changes sign at these points.

$$${f{''}}{\left({x}\right)}>{0}$$$ on $$${\left(-\infty,-{1}\right)}\cup{\left({0.27},{3.73}\right)}$$$ and $$${f{''}}{\left({x}\right)}<{0}$$$ on $$${\left(-{1},{0.27}\right)}\cup{\left({3.73},\infty\right)}$$$.

So we have following "important" points:

Function is increasing on $$${\left(-\infty,-{0.41}\right)}\cup{\left({2.41},\infty\right)}$$$ and decreasing on $$${\left(-{0.41},{2.41}\right)}$$$.

Function is concave upward on $$${\left(-\infty,-{1}\right)}\cup{\left({0.27},{3.73}\right)}$$$ and concave downward on $$${\left(-{1},{0.27}\right)}\cup{\left({3.73},\infty\right)}$$$.

Horizontal asymptote is $$${y}={1}$$$.

Example 4. Sketch graph of the function $$${f{{\left({x}\right)}}}=\frac{{{\left({x}-{1}\right)}}^{{3}}}{{{\left({x}+{1}\right)}}^{{2}}}$$$.

There is no simpler function that initial function is obtained from.

Function is neither even nor odd and not periodic.

$$${f{{\left({0}\right)}}}=\frac{{{\left({0}-{1}\right)}}^{{3}}}{{{\left({0}+{1}\right)}}^{{2}}}=-{1}$$$, so y-intercept is -1.

x-intercepts are points where $$${f{{\left({x}\right)}}}={0}$$$: $$${{\left({x}-{1}\right)}}^{{3}}={0}$$$ which gives $$${x}={1}$$$.

Vertical asymptote is $$${x}=-{1}$$$ because $$$\lim_{{{x}\to-{{1}}^{+}}}\frac{{{\left({x}-{1}\right)}}^{{3}}}{{{\left({x}+{1}\right)}}^{{2}}}=-\infty$$$.

There are no horizontal asymptotes because $$$\lim_{{{x}\to\infty}}\frac{{{\left({x}-{1}\right)}}^{{3}}}{{{\left({x}+{1}\right)}}^{{2}}}=\infty$$$ and $$$\lim_{{{x}\to-\infty}}\frac{{{\left({x}-{1}\right)}}^{{3}}}{{{\left({x}+{1}\right)}}^{{2}}}=-\infty$$$.

Since $$$\lim_{{{x}\to\infty}}\frac{{{f{{\left({x}\right)}}}}}{{x}}=\lim_{{{x}\to\infty}}\frac{{{{x}}^{{3}}-{3}{{x}}^{{2}}+{3}{x}-{1}}}{{{{x}}^{{3}}+{2}{{x}}^{{2}}+{x}}}={1}$$$ then $$${m}={1}$$$. Now $$$\lim_{{{x}\to\infty}}{\left({f{{\left({x}\right)}}}-{m}{x}\right)}=\frac{{{{x}}^{{3}}-{3}{{x}}^{{2}}+{3}{x}-{1}}}{{{{x}}^{{2}}+{2}{x}+{1}}}-{x}=\lim_{{{x}\to\infty}}\frac{{-{5}{{x}}^{{2}}+{2}{x}-{1}}}{{{{x}}^{{2}}-{2}{x}+{1}}}=-{5}$$$.

So, there is slant asymptote $$${y}={x}-{5}$$$.

We have that $$${f{'}}{\left({x}\right)}=\frac{{{\left({{\left({x}-{1}\right)}}^{{3}}\right)}'{{\left({x}+{1}\right)}}^{{2}}-{{\left({x}-{1}\right)}}^{{3}}{\left({{\left({x}+{1}\right)}}^{{2}}\right)}'}}{{{\left({x}+{1}\right)}}^{{4}}}=\frac{{{{\left({x}-{1}\right)}}^{{2}}{\left({x}+{5}\right)}}}{{{\left({x}+{1}\right)}}^{{3}}}$$$.

Derivative equals 0 when $$${x}={1},{x}=-{5}$$$ and doesn't exist when $$${x}=-{1}$$$.

Since $$${{\left({x}-{1}\right)}}^{{2}}\ge{0}$$$ then these factor doesn't influence sign of derivative.

Thus $$${f{'}}{\left({x}\right)}>{0}$$$ when $$${x}\in{\left(-\infty,-{5}\right)}\cup{\left(-{1},\infty\right)}$$$ and $$${f{'}}{\left({x}\right)}<{0}$$$ when $$${x}\in{\left(-{5},-{1}\right)}$$$.

According to First Derivative Test $$${x}=-{5}$$$ is local maximum and $$${x}={1}$$$ is not an extremum.

Point $$${x}=-{1}$$$ is not in the domain of function, so clearly not an extremum.

Second derivative is $$${f{''}}{\left({x}\right)}={24}\frac{{{x}-{1}}}{{{\left({x}+{1}\right)}}^{{4}}}$$$.

So, point where second derivative equals 0 is $$${x}={1}$$$.

Clearly it is inflection point because second derivative changes sign at this point.

$$${f{''}}{\left({x}\right)}>{0}$$$ when $$${x}>{1}$$$ and $$${f{''}}{\left({x}\right)}<{0}$$$ when $$${x}<{1}$$$.

Let's add a couple of control points:

$$${x}={3}$$$, $$${f{{\left({3}\right)}}}={0.5}$$$.

$$${x}=-{8}$$$, $$${f{{\left(-{8}\right)}}}\approx-{14.88}$$$.

So, we have following points:

Function is increasing on $$${\left(-\infty,-{5}\right)}\cup{\left(-{1},\infty\right)}$$$ and decreasing on $$${\left(-{5},-{1}\right)}$$$.

Function is concave upward on $$${\left({1},\infty\right)}$$$ and concave downward on $$${\left(-\infty,{1}\right)}$$$.

Vertical asymptote is $$${x}=-{1}$$$ and slant asymptote is $$${y}={x}-{5}$$$.