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Precalculus : Graph a Polynomial Function

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Example Questions

Example Question #1 : Graph A Polynomial Function

Which of the following is an accurate graph of?

Possible Answers:

Varsity10

Varsity12

Varsity2

Varsity11

Varsity1

Correct answer:

Varsity1

Explanation:

f(x) is a parabola, because of the general $f(x) = x^{2}$ structure. The parabola opens downward because f(x) < 0 .

Solving $-(x + 4)^{2} = 0$ tells the x-value of the x-axis intercept;

$0 = (x + 4)^{2}$

$\sqrt{0} = \sqrt{(x + 4)^{2}}$

0 = x + 4

-4 = x

The resulting x-axis intercept is: (-4, 0) .

Setting x = 0 tells the y-value of the y-axis intercept;

= -1(16)

= -16

The resulting y-axis intercept is: (0, -16)

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Example Question #3 : Graphs Of Polynomial Functions

Give the -intercept of the graph of the function

$f(x) = 5 \cdot 4^{x- 3}- 3$

Round to the nearest tenth, if applicable.

Possible Answers:

(2.63,0)

(3.65,0)

( -3.08, 0)

( -2.92, 0)

The graph has no -interceptx

Correct answer:

(2.63,0)

Explanation:

The -intercept is (a,0) , where f(a)= 0 :

$f(x) = 5 \cdot 4^{x- 3}- 3$

$5 \cdot 4^{a- 3}- 3 = 0$

$5 \cdot 4^{a- 3}- 3 + 3= 0 + 3$

$5 \cdot 4^{a- 3}= 3$

$5 \cdot 4^{a- 3} \div 5= 3 \div 5$

$4^{a- 3} = 0.6$

$\ln \left (4^{a- 3} \right )=\ln 0.6$

$(a- 3)\ln 4 =\ln 0.6$

$a- 3=\frac{\ln 0.6}{\ln 4 } \approx \frac{-0.5108}{1.3863}\approx -0.3685$

$a- 3+3 \approx -0.3685 + 3$

$a \approx 2.63$

The -intercept is (2.63,0) .

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Example Question #1 : Graph Polynomial Functions, Identify Zeros, Factor, And Identify End Behavior.: Css.Math.Content.Hsf If.C.7c

Graph the following function and identify the zeros.

x^3-2x^2-x+2

Possible Answers:

Screen shot 2016 01 13 at 9.55.24 am

$\textup{Roots: }x=-1, x=1, x=2$

Screen shot 2016 01 13 at 9.50.10 am

$\textup{Roots: } x=1, x=2$

Screen shot 2016 01 13 at 12.16.31 pm

$\textup{Roots: }x=-1, x=0, x=2$

Screen shot 2016 01 13 at 12.17.10 pm

$\textup{Roots: }x=-1, x=2$

Screen shot 2016 01 13 at 12.16.52 pm

$\textup{Roots: }x=-1, x=1, x=-2$

Correct answer:

Screen shot 2016 01 13 at 9.55.24 am

$\textup{Roots: }x=-1, x=1, x=2$

Explanation:

This question tests one's ability to graph a polynomial function.

For the purpose of Common Core Standards, "graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior" falls within the Cluster C of "Analyze Functions Using Different Representations" concept (CCSS.MATH.CONTENT.HSF-IF.C.7).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic technique to factor the function.

x^3-2x^2-x+2

Separating the function into two parts...

(x^3-2x^2)+(-x+2)

Factoring a negative one from the second set results in...

(x^3-2x^2)-(x-2)

Factoring out x^2 from the first set results in...

x^2(x-2)-1(x-2)

The new factored form of the function is,

(x^2-1)(x-2) .

Now, recognize that the first binomial is a perfect square for which the following formula can be used

$(x^2-b^2)=(x-\sqrt{b})(x+\sqrt{b})$

since

$b=1\Rightarrow \sqrt{b}=\pm1$

thus the simplified, factored form is,

(x-1)(x+1)(x-2) .

Step 2: Identify the roots of the function.

To find the roots of a function set its factored form equal to zero and solve for the possible x values.

$\\(x-1)(x+1)(x-2)=0 \\x-1=0\Rightarrow x=1 \\x+1=0\Rightarrow x=-1 \\x-2=0\Rightarrow x=2$

Step 3: Create a table of (x,y) pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \\ \hline -2& f(-2) & -12 \\ -1 & f(-1) & 0 \\ 0&f(0)& 2\\ 1.5&f(1.5)&-0.625\\2&f(2)&0 \\ 3&f(3)&8 \\\hline \end{tabular}$

The values in the table are found by substituting in the x values into the function as follows.

$\\f(-2)=(-2)^3-2(-2)^2-(-2)+2=-8-8+2+2=-16+4=-12 \\f(-1)=(-1)^3-2(-1)^2-(-1)+2=-1-2+1+2=0 \\f(0)=(0)^3-2(0)^2-(0)+2=0-0-0+2=2 \\f(1.5)=(1.5)^3-2(1.5)^2-(1.5)+2=-0.625 \\f(2)=(2)^3-2(2)^2-(2)+2=8-8-2+2=0 \\f(3)=(3)^3-2(3)^2-(3)+2=27-18-3+2=8$

Step 4: Plot the points on a coordinate grid and connect them with a smooth curve.

Screen shot 2016 01 13 at 9.55.24 am

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Example Question #2 : Graph A Polynomial Function

Graph the function and identify the roots.

100x^2-1

Possible Answers:

Question12

$\text{Roots: }x=\pm\frac{1}{10}$

Question2

$\text{Roots: }x=\pm\frac{1}{10}$

Question3

$\text{Roots: }x=\pm3$

Question5

$\text{Roots: }x=\pm5$

Question6

$\text{Roots: }x=\pm\frac{1}{10}$

Correct answer:

Question12

$\text{Roots: }x=\pm\frac{1}{10}$

Explanation:

This question tests one's ability to graph a polynomial function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic technique to factor the function.

Recognize that the binomial is a perfect square for which the following formula can be used

$(a^2x^2-b^2)=(\sqrt{a}x-\sqrt{b})(\sqrt{a}x+\sqrt{b})$

since

$\\a=100\Rightarrow \sqrt{a}=\pm10 \\b=1\Rightarrow \sqrt{b}=\pm1$

thus the simplified, factored form is,

(10x-1)(10x+1) .

Step 2: Identify the roots of the function.

To find the roots of a function set its factored form equal to zero and solve for the possible x values.

$\\10x-1=0 \\10x=1\\ \\x=\frac{1}{10}$ $\\10x+1=0 \\10x=-1 \\\\x=-\frac{1}{10}$

Step 3: Create a table of (x,y) pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \\ \hline -1& f(-1) & 99 \\ 0 & f(0) & -1 \\ 1&f(1)& 99 \\\hline \end{tabular}$

The values in the table are found by substituting in the x values into the function as follows.

$\\f(-1)=100(-1)^2-1=100-1=99 \\f(0)=100(0)^2-1=0-1=-1 \\f(1)=100(1)^2-1=100-1=99$

Step 4: Plot the points on a coordinate grid and connect them with a smooth curve.

Question12

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Example Question #3 : Graph A Polynomial Function

Graph the function and identify its roots.

9x^2-16

Possible Answers:

Question5

$\text{Roots: }x=\pm\frac{3}{4}$

Question3

$\text{Roots: }x=\pm\frac{3}{4}$

Screen shot 2016 01 13 at 12.16.31 pm

$\text{Roots: }x=\pm\frac{1}{4}$

Question4

$\text{Roots: }x=\pm\frac{4}{3}$

Question6

$\text{Roots: }x=\pm\frac{4}{3}$

Correct answer:

Question6

$\text{Roots: }x=\pm\frac{4}{3}$

Explanation:

This question tests one's ability to graph a polynomial function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic technique to factor the function.

Recognize that the binomial is a perfect square for which the following formula can be used

$(a^2x^2-b^2)=(\sqrt{a}x-\sqrt{b})(\sqrt{a}x+\sqrt{b})$

since

$\\a=9\Rightarrow \sqrt{a}=\pm3 \\b=16\Rightarrow \sqrt{b}=\pm4$

thus the simplified, factored form is,

(3x-4)(3x+4) .

Step 2: Identify the roots of the function.

To find the roots of a function set its factored form equal to zero and solve for the possible x values.

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

Step 3: Create a table of (x,y) pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \\ \hline -2& f(-2) & 20 \\ -4/3 & f(-4/3) & 0 \\ 0&f(0)& -16\\ 1&f(1)&-7\\\hline \end{tabular}$

The values in the table are found by substituting in the x values into the function as follows.

$\\f(-2)=9(-2)^2-16=9(4)-16=36-16=20 \\f(-4/3)=9(-4/3)^2-16=16-16=0 \\f(0)=9(0)^2-10-16=-16 \\f(1)=9(1)^2-16=9-16=-7$

Step 4: Plot the points on a coordinate grid and connect them with a smooth curve.

Question6

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Example Question #1 : Graph Polynomial Functions, Identify Zeros, Factor, And Identify End Behavior.: Css.Math.Content.Hsf If.C.7c

Graph the function and identify its roots.

4x^2-36

Possible Answers:

Screen shot 2016 01 13 at 12.16.31 pm

$\text{Roots: }x=\pm3$

Question4

$\text{Roots: }x=\pm3$

Question2

$\text{Roots: }x=\pm2$

Question3

$\text{Roots: }x=\pm3$

Screen shot 2016 01 13 at 12.16.52 pm

$\text{Roots: }x=-1, x=2$

Correct answer:

Question4

$\text{Roots: }x=\pm3$

Explanation:

This question tests one's ability to graph a polynomial function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic technique to factor the function.

Recognize that the binomial is a perfect square for which the following formula can be used

$(a^2x^2-b^2)=(\sqrt{a}x-\sqrt{b})(\sqrt{a}x+\sqrt{b})$

since

$\\a=4\Rightarrow \sqrt{a}=\pm2 \\b=36\Rightarrow \sqrt{b}=\pm6$

thus the simplified, factored form is,

(2x-6)(2x+6) .

Step 2: Identify the roots of the function.

To find the roots of a function set its factored form equal to zero and solve for the possible x values.

$\\2x-6=0 \\2x=6 \\x=3$ $\\2x+6=0 \\2x=-6 \\x=-3$

Step 3: Create a table of (x,y) pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \\ \hline -3& f(-3) & 0 \\ -1 & f(-1) & -32 \\ 0&f(0)& -36\\ 1&f(1)&-32\\3&f(3)&0 \\\hline \end{tabular}$

The values in the table are found by substituting in the x values into the function as follows.

$\\f(-3)=4(-3)^2-36=4(9)-36=36-36=0 \\f(-1)=4(-1)^2-36=4-36=-32 \\f(0)=4(0)^2-36=0-36=-36 \\f(1)=4(1)^2-36=4-36=-32 \\f(3)=4(3)^2-36=4(9)-36=36-36=0$

Step 4: Plot the points on a coordinate grid and connect them with a smooth curve.

Question4

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Precalculus : Graph a Polynomial Function

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #1 : Graph A Polynomial Function

Example Question #3 : Graphs Of Polynomial Functions

Example Question #1 : Graph Polynomial Functions, Identify Zeros, Factor, And Identify End Behavior.: Css.Math.Content.Hsf If.C.7c

Example Question #2 : Graph A Polynomial Function

Example Question #3 : Graph A Polynomial Function

Example Question #1 : Graph Polynomial Functions, Identify Zeros, Factor, And Identify End Behavior.: Css.Math.Content.Hsf If.C.7c

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