Graphing Polynomials

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Question 1 of 3

1. Question

Graph

y = \frac{4 x - 10}{x - 3}

$y=(4x-10)/(x-3)$

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Division of Polynomials

$\mathsf{\frac{P}{Divisor}=Quotient+\frac{R}{Divisor}}$

where

$\mathsf{P}$ is the Polynomial and

$\mathsf{R}$ is the Remainder

Long Division

Use long division when a polynomial is divided by a binomial

General Form of a Hyperbola

$y=a/(x-h)+k$

Asymptotes:

$y=k$ and

$x=h$

This polynomial can be written in the General Form of a Hyperbola and then graphed as a hyperbola.

First, divide the fraction using Long Division.

Substitute components into the formula

$\mathsf{P}$ (Polynomial)	$=$	$4x-10$
$\mathsf{Divisor}$	$=$	$x-3$

$=$

Next, solve for the quotient

First term of the quotient:

Divide the first term of the Polynomial by the first term of the Divisor. Place this above the Polynomial

$4xdividex$

$=$

$4$

Multiply $4$ to the divisor. Place this under the Polynomial

$4$

$(x-3)$

$=$

$4x-12$

Subtract $4x-12$ and write the difference one line below

Since

$2$ cannot be divided by the divisor anymore, it is left as a remainder

This also means that the expression at the very top is the Quotient

Combine and substitute the components into the Division of Polynomials formula

$\mathsf{P}$ (Polynomial)	$=$	$4x-10$
$\mathsf{Divisor}$	$=$	$x-3$
$\mathsf{Quotient}$	$=$	$4$
$\mathsf{Remainder}$	$=$	$2$

$\mathsf{\frac{P}{Divisor}}$	$=$	$\mathsf{Quotient+\frac{R}{Divisor}}$	Division of Polynomials

$\frac{4x-10}{x-3}$	$=$	$4+\frac{2}{x-3}$	Substitute

$y$	$=$	$\frac{2}{x-3}+4$

Notice that this expression is in the General Form of a Hyperbola and hence can be graphed as a hyperbola

Identify components of the Hyperbola

$y$	$=$	$a/(x-h)+k$

$y$	$=$	$2/(x-3)+4$

$a$	$=$	$2$
$h$	$=$	$3$
$k$	$=$	$4$

Graph the asymptotes

$x=h$ becomes

$x=3$

$y=k$ becomes

$y=4$

Use sample points to help on graphing the curves

First point using $x=2$ :

$y$	$=$	$2/(x-3)+4$

$y$	$=$	$2/(2-3)+4$	Substitute $x=2$

$y$	$=$	$2/(-1)+4$

$y$	$=$	$-2+4$
$y$	$=$	$2$

Second point using $x=4$ :

$y$	$=$	$2/(x-3)+4$

$y$	$=$	$2/(4-3)+4$	Substitute $x=4$

$y$	$=$	$2/(1)+4$

$y$	$=$	$2+4$
$y$	$=$	$6$

Plot the points $(2,2)$ and $(4,6)$

Finally, graph the curves over the sample points. Make sure that these curves come close but never intersect the asymptotes.

Question 2 of 3

2. Question

Graph

$y=(2+3x)/(x-1)$

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Chapters

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Division of Polynomials

$\mathsf{\frac{P}{Divisor}=Quotient+\frac{R}{Divisor}}$

where

$\mathsf{P}$ is the Polynomial and

$\mathsf{R}$ is the Remainder

Long Division

Use long division when a polynomial is divided by a binomial

General Form of a Hyperbola

$y=a/(x-h)+k$

Asymptotes:

$y=k$ and

$x=h$

This polynomial can be written in the General Form of a Hyperbola and then graphed as a hyperbola.

First, divide the fraction using Long Division.

Arrange the terms in descending order of powers, then substitute components into the formula

$\mathsf{P}$ (Polynomial)	$=$	$2+3x$
	$=$	$3x+2$
$\mathsf{Divisor}$	$=$	$x-1$

$=$

Next, solve for the quotient

First term of the quotient:

Divide the first term of the Polynomial by the first term of the Divisor. Place this above the Polynomial

$3xdividex$

$=$

$3$

Multiply $3$ to the divisor. Place this under the Polynomial

$3$

$(x-1)$

$=$

$3x-3$

Subtract $3x-3$ and write the difference one line below

Since

$5$ cannot be divided by the divisor anymore, it is left as a remainder

This also means that the expression at the very top is the Quotient

Combine and substitute the components into the Division of Polynomials formula

$\mathsf{P}$ (Polynomial)	$=$	$3x+2$
$\mathsf{Divisor}$	$=$	$x-1$
$\mathsf{Quotient}$	$=$	$3$
$\mathsf{Remainder}$	$=$	$5$

$\mathsf{\frac{P}{Divisor}}$	$=$	$\mathsf{Quotient+\frac{R}{Divisor}}$	Division of Polynomials

$\frac{3x+2}{x-1}$	$=$	$3+\frac{5}{x-1}$	Substitute

$y$	$=$	$\frac{5}{x-1}+3$

Notice that this expression is in the General Form of a Hyperbola and hence can be graphed as a hyperbola

Identify components of the Hyperbola

$y$	$=$	$a/(x-h)+k$

$y$	$=$	$5/(x-1)+3$

$a$	$=$	$5$
$h$	$=$	$1$
$k$	$=$	$3$

Graph the asymptotes

$x=h$ becomes

$x=1$

$y=k$ becomes

$y=3$

Use sample points to help on graphing the curves

First point using $x=0$ :

$y$	$=$	$5/(x-1)+3$

$y$	$=$	$5/(0-1)+3$	Substitute $x=0$

$y$	$=$	$5/(-1)+3$

$y$	$=$	$-5+3$
$y$	$=$	$-2$

Second point using $x=2$ :

$y$	$=$	$5/(x-1)+3$

$y$	$=$	$5/(2-1)+3$	Substitute $x=2$

$y$	$=$	$5/(1)+3$

$y$	$=$	$5+3$
$y$	$=$	$8$

Plot the points $(0,-2)$ and $(2,8)$

Finally, graph the curves over the sample points. Make sure that these curves come close but never intersect the asymptotes.

Question 3 of 3

3. Question

Graph

$P(x)=x^3-2x^2-19x+20$

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Long Division

Use long division when a polynomial is divided by a binomial

Factor Theorem

factor

$(P(x))/(x-a)$ only if

$P(a)=0$

Since the polynomial is in cubic form, we can graph it by finding its zeroes then sketching the curve along.

First, factor out the polynomial completely using Factor Theorem

Try out numbers that divide to the constant term which is

$20$ as the value of

$a$ then substitute into the polynomial

$20$ can be divided by:

$+-$ $1,2,4,5,10,20$

Trying out $a=2$ :

$P(x)$	$=$	$x^3-2x^2-19x+20$
$P(a)$	$=$	$a^3-2a^2-19a+20$	Replace $x$ with $a$
$P(2)$	$=$	$2^3-2(2)^2-19(2)+20$	Substitute $a=2$
	$=$	$8-2(4)-38+20$
	$=$	$8-8-38+20$
	$=$	$-18$

Since

$P(2)≠0$ ,

$(x-2)$ is not a factor of the polynomial

Trying out $a=1$ :

$P(x)$	$=$	$x^3-2x^2-19x+20$
$P(a)$	$=$	$a^3-2a^2-19a+20$	Replace $x$ with $a$
$P(1)$	$=$	$1^3-2(1)^2-19(1)+20$	Substitute $a=1$
	$=$	$1-2(1)-19+20$
	$=$	$1-2-19+20$
	$=$	$0$