New Notes Blog: `y=(x+4)/(x-3)` Graph the function. State the domain and range.

Thursday, 30 April 2015

$\displaystyle{y}=\frac{{{x}+{4}}}{{{x}-{3}}}$ Graph the function. State the domain and range.

To be able to graph the rational function $\displaystyle{y}=\frac{{{x}+{4}}}{{{x}-{3}}}$ , we solve for possible asymptotes.

Vertical asymptote exists at $\displaystyle{x}={a}$ that will satisfy $\displaystyle{D}{\left({x}\right)}={0}$ on a rational function $\displaystyle{f{{\left({x}\right)}}}=\frac{{{N}{\left({x}\right)}}}{{{D}{\left({x}\right)}}}$ . To solve for the vertical asymptote, we equate the expression at denominator side to $\displaystyle{0}$ and solve for x.

In $\displaystyle{y}=\frac{{{x}+{4}}}{{{x}-{3}}}$ , the $\displaystyle{D}{\left({x}\right)}={x}-{3}.$

Then, $\displaystyle{D}{\left({x}\right)}={0}$ will be:

$\displaystyle{x}-{3}={0}$

$\displaystyle{x}-{3}+{3}={0}+{3}$

$\displaystyle{x}={3}$

The vertical asymptote exists at $\displaystyle{x}={3}$ .

To determine the horizontal asymptote for a given function: $\displaystyle{f{{\left({x}\right)}}}=\frac{{{a}{{x}}^{{n}}+\ldots}}{{{b}{{x}}^{{m}}+\ldots}},$ we follow the conditions:

when $\displaystyle{n}\lt{m}$ horizontal asymptote: $\displaystyle{y}={0}$

$\displaystyle{n}={m}$ horizontal asymptote: $\displaystyle{y}=\frac{{a}}{{b}}$

$\displaystyle{n}\gt{m}$ horizontal asymptote: NONE

In $\displaystyle{y}=\frac{{{x}+{4}}}{{{x}-{3}}}$ , the leading terms are $\displaystyle{a}{{x}}^{{n}}={x}{\quad\text{or}\quad}{1}{{x}}^{{1}}$ and $\displaystyle{b}{{x}}^{{m}}={x}{\quad\text{or}\quad}{1}{{x}}^{{1}}$ . The values $\displaystyle{n}={1}$ and $\displaystyle{m}={1}$ satisfy the condition: n=m. Then, horizontal asymptote exists at $\displaystyle{y}=\frac{{1}}{{1}}$ or $\displaystyle{y}={1}$ .

To solve for possible y-intercept, we plug-in $\displaystyle{x}={0}$ and solve for $\displaystyle{y}$ .

$\displaystyle{y}=\frac{{{0}+{4}}}{{{0}-{3}}}$

$\displaystyle{y}=\frac{{4}}{{-{3}}}$

$\displaystyle{y}=-\frac{{4}}{{3}}{\quad\text{or}\quad}-{1.333}$ (approximated value)

Then, y-intercept is located at a point $\displaystyle{\left({0},-{1.333}\right)}$ .

To solve for possible x-intercept, we plug-in $\displaystyle{y}={0}$ and solve for $\displaystyle{x}$ .

$\displaystyle{0}=\frac{{{x}+{4}}}{{{x}-{3}}}$

$\displaystyle{0}\cdot{\left({x}-{3}\right)}=\frac{{{x}+{4}}}{{{x}-{3}}}\cdot{\left({x}-{3}\right)}$

$\displaystyle{0}={x}+{4}$

$\displaystyle{0}-{4}={x}+{4}-{4}$

$\displaystyle-{4}={x}{\quad\text{or}\quad}{x}=-{4}$

Then, x-intercept is located at a point $\displaystyle{\left(-{4},{0}\right)}$ .

Solve for additional points as needed to sketch the graph.

When $\displaystyle{x}={2}$ , the $\displaystyle{y}=\frac{{{2}+{4}}}{{{2}-{3}}}=\frac{{6}}{{-{1}}}=-{6}$ . point: $\displaystyle{\left({2},-{6}\right)}$

When $\displaystyle{x}={4}$ , the $\displaystyle{y}=\frac{{{4}+{4}}}{{{4}-{3}}}=\frac{{8}}{{1}}={8}$ . point: $\displaystyle{\left({4},{8}\right)}$

When $\displaystyle{x}={10}$ , the $\displaystyle{y}=\frac{{{10}+{4}}}{{{10}-{3}}}=\frac{{14}}{{7}}={2}$ . point: $\displaystyle{\left({10},{2}\right)}$

When $\displaystyle{x}=-{16}$ , the $\displaystyle{y}=\frac{{-{16}+{4}}}{{-{16}-{3}}}=-\frac{{12}}{{-{19}}}\approx{0.632}$ . point: $\displaystyle{\left(-{16},{0.632}\right)}$

Applying the listed properties of the function, we plot the graph as:

You may check the attached file to verify the plot of asymptotes and points.

As shown on the graph, the domain: $\displaystyle{\left(-\infty,{3}\right)}\cup{\left({3},\infty\right)}$

and range: $\displaystyle{\left(-\infty,{1}\right)}\cup{\left({1},\infty\right)}.$

The domain of the function is based on the possible values of $\displaystyle{x}.$ The $\displaystyle{x}={3}$ excluded due to the vertical asymptote.

The range of the function is based on the possible values of $\displaystyle{y}$ . The $\displaystyle{y}={1}$ is excluded due to the horizontal asymptote.

New Notes Blog

Thursday, 30 April 2015

$\displaystyle{y}=\frac{{{x}+{4}}}{{{x}-{3}}}$ Graph the function. State the domain and range.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Thursday, 30 April 2015

Graph the function. State the domain and range.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{y}=\frac{{{x}+{4}}}{{{x}-{3}}}$ Graph the function. State the domain and range.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?