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Graphing Polynomials

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Graphing Polynomials

Graphing Polynomials

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Question 1 of 3

1. Question

Graph

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

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Chapters

Chapters

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

Chapters

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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Chapters

Chapters

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 $a=1$
	$=$	$1-2(1)-19+20$
	$=$	$1-2-19+20$
	$=$	$0$

Since

$P(1)=0$ , $(x-1)$ is the first factor of the polynomial

Find the second factor using Long Division.

Substitute components into the formula

$\mathsf{P}$ (Polynomial)	$=$	$x^3-2x^2-19x+20$
$\mathsf{Divisor}$	$=$	$x-1$

$=$

Next, solve for each term of 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

$x^3dividex$

$=$

$x^2$

Multiply $x^2$ to the divisor. Place this under the Polynomial

$x^2$

$(x-1)$

$=$

$x^3-x^2$

Subtract $x^3-x^2$ and write the difference one line below

Drop down

$-19x$ and repeat the process to get the second term of the quotient

Second term of the quotient:

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

$-x^2dividex$

$=$

$-x$

Multiply $-x$ to the divisor. Place this one line below

$-x$

$(x-1)$

$=$

$-x^2+x$

Subtract $-x^2+x$ and write the difference one line below

Drop down

$20$ and repeat the process to get the third term of the quotient

Third term of the quotient:

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

$-20xdividex$

$=$

$-20$

Multiply $-20$ to the divisor. Place this under the Polynomial

$-20$

$(x-1)$

$=$

$-20x+20$

Subtract $-20x+20$ and write the difference one line below

Since

$r=0$ and cannot be divided anymore, the quotient is

$x^2-x-20$

So far, we have factored the polynomial as $(x-1)$

$(x^2-x-20)$

Factor out the polynomial further by applying cross method

$(x-1)(x^2-x-20)$

$(x-1)(x+4)(x-5)$

Notice that in this form, we can easily get the zeroes of the cubic curve.

Find the zeroes of the curve by equating the factorised polynomial to

$0$ and solving each value of

$x$

$(x-1)(x+4)(x-5)$

$=$

$0$

First

$x$ value:

$x-1$	$=$	$0$
$x-1$ $+1$	$=$	$0$ $+1$	Add $1$ to both sides
$x$	$=$	$1$

Second

$x$ value:

$x+4$	$=$	$0$
$x+4$ $-4$	$=$	$0$ $-4$	Subtract $4$ from both sides
$x$	$=$	$-4$

Third

$x$ value:

$x-5$	$=$	$0$
$x-5$ $+5$	$=$	$0$ $+5$	Add $5$ to both sides
$x$	$=$	$5$

Hence, the zeroes will be at

$x=1,-4,5$

Finally, graph the cubic curve over the zeroes. Note that since the coefficient of the

$x^3$ term is positive, the curve should start from the bottom left

Quizzes

Division of Polynomials
Remainder Theorem
Factor Theorem
Graphing Polynomials

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