Notes: `h(x)=(5x+3)/(-x+16)` Graph the function. State the domain and range.

Wednesday, 2 December 2015

$\displaystyle{h}{\left({x}\right)}=\frac{{{5}{x}+{3}}}{{-{x}+{16}}}$ Graph the function. State the domain and range.

To be able to graph the rational function $\displaystyle{y}=\frac{{{5}{x}+{3}}}{{-{x}+{16}}}$ , 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 $\displaystyle{x}$ .

In $\displaystyle{y}=\frac{{{5}{x}+{3}}}{{-{x}+{16}}}$ , the $\displaystyle{D}{\left({x}\right)}=-{x}+{16}$ .

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

$\displaystyle-{x}+{16}={0}$

$\displaystyle{x}={16}$

The vertical asymptote exists at $\displaystyle{x}={16}$ .

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 $\displaystyle{y}=\frac{{{5}{x}+{3}}}{{-{x}+{16}}}$ , the leading terms are $\displaystyle{a}{{x}}^{{n}}={5}{x}{\quad\text{or}\quad}{5}{{x}}^{{1}}$ 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 $\displaystyle{y}=\frac{{5}}{{-{1}}}$ or $\displaystyle{y}=-{5}$ .

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

$\displaystyle{y}=\frac{{{5}\cdot{0}+{3}}}{{-{0}+{16}}}$

$\displaystyle{y}=\frac{{{0}+{3}}}{{{0}+{16}}}$

$\displaystyle{y}=\frac{{3}}{{16}}{\quad\text{or}\quad}{0.188}$ (approximated value)

Then, y-intercept is located at a point $\displaystyle{\left({0},{0.188}\right)}$ .

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

$\displaystyle{0}=\frac{{{5}{x}+{3}}}{{-{x}+{16}}}$

$\displaystyle{0}\cdot{\left(-{x}+{16}\right)}=\frac{{{5}{x}+{3}}}{{-{x}+{16}}}\cdot{\left(-{x}+{16}\right)}$

$\displaystyle{0}={5}{x}+{3}$

$\displaystyle-{3}={5}{x}$

$\displaystyle{x}=\frac{{-{3}}}{{5}}=-{0.6}$

Then, x-intercept is located at a point $\displaystyle{\left(-{0.6},{0}\right)}$ .

Solve for additional points as needed to sketch the graph.

When $\displaystyle{x}={11}$ , the $\displaystyle{y}=\frac{{{5}\cdot{11}+{3}}}{{-{11}+{16}}}=\frac{{58}}{{5}}={11.6}$ . point: $\displaystyle{\left({11},{11.6}\right)}$

When $\displaystyle{x}={20}$ , the $\displaystyle{y}=\frac{{{5}\cdot{20}+{3}}}{{-{20}+{16}}}=\frac{{103}}{{-{4}}}=-{25.75}$ . point: $\displaystyle{\left({20},-{25.75}\right)}$

When $\displaystyle{x}={30}$ , the $\displaystyle{y}=\frac{{{5}\cdot{30}+{3}}}{{-{30}+{16}}}=\frac{{153}}{{-{14}}}\approx-{11}$ . point: $\displaystyle{\left({30},-{11}\right)}$

When $\displaystyle{x}=-{26}$ , the y $\displaystyle=\frac{{{5}{\left(-{26}\right)}+{3}}}{{-{\left(-{26}\right)}+{16}}}=-\frac{{127}}{{42}}\approx-{3.024}$ . point: $\displaystyle{\left(-{26},-{3.024}\right)}$

As shown on the graph attached below, the domain: $\displaystyle{\left(-\infty,{16}\right)}\cup{\left({16},\infty\right)}$

and range: $\displaystyle{\left(-\infty,-{5}\right)}\cup{\left(-{5},\infty\right)}.$

The domain of the function is based on the possible values of x. The $\displaystyle{x}={16}$ excluded due to the vertical asymptote.

The range of the function is based on the possible values of y. The $\displaystyle{y}=-{5}$ is excluded due to the horizontal asymptote.

Notes

Wednesday, 2 December 2015

$\displaystyle{h}{\left({x}\right)}=\frac{{{5}{x}+{3}}}{{-{x}+{16}}}$ Graph the function. State the domain and range.

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Wednesday, 2 December 2015

Graph the function. State the domain and range.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{h}{\left({x}\right)}=\frac{{{5}{x}+{3}}}{{-{x}+{16}}}$ Graph the function. State the domain and range.

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?