Graph Polynomial Functions, Identify Zeros, Factor, and Identify End Behavior.: CSS.Math.Content.HSF-IF.C.7c

Help Questions

Common Core: High School - Functions › Graph Polynomial Functions, Identify Zeros, Factor, and Identify End Behavior.: CSS.Math.Content.HSF-IF.C.7c

Questions 1 - 10

Graph the function and identify the roots.

Question11

$\text{Roots: }x=\pm\frac{5}{2}$

Question2

$\text{Roots: }x=\pm\frac{5}{2}$

Question3

$\text{Roots: }x=\pm\frac{5}{2}$

Question4

$\text{Roots: }x=\pm\frac{5}{2}$

Question5

$\text{Roots: }x=\pm{5}$

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.

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

$\a=4\Rightarrow \sqrt{a}=\pm2 \b=25\Rightarrow \sqrt{b}=\pm5$

thus the simplified, factored form is,

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.

$\2x-5=0 \2x=5 \\x=\frac{5}{2}$ $\2x+5=0 \2x=-5 \\x=-\frac{5}{2}$

Step 3: Create a table of pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \ \hline -3& f(-3) & 11 \ -5/2 & f(-5/2) & 0 \ 0&f(0)& -25\ 5/2&f(5/2)&0\3&f(3)&11 \\hline \end{tabular}$

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

$\f(-3)=4(-3)^2-25=4(9)-35=36-25=11 \f(-5/2)=4(-5/2)^2-25=4(25/4)-25=0 \f(0)=4(0)^2-25=0-25=-25 \f(5/2)=4(5/2)^2-25=4(25/4)-25=0 \f(3)=4(3)^2-25=4(9)-25=36-25=11$

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

Question11

Graph the function and identify the roots.

Question3

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

Question4

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

Question2

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

Screen shot 2016 01 13 at 12.16.52 pm

$\text{Root: }x=-1$

Screen shot 2016 01 13 at 9.50.10 am

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

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

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

since

$b=9\Rightarrow \sqrt{b}=\pm3$

thus the simplified, factored form is,

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.

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

Step 3: Create a table of pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \ \hline -3& f(-3) & 0 \ -1 & f(-1) & -8 \ 0&f(0)& -9\ 1&f(1)&-8\3&f(3)&0 \\hline \end{tabular}$

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

$\f(-3)=x^2-9=(-3)^2-9=9-9=0 \f(-1)=(-1)^2-9=1-9=-8 \f(0)=0^2-9=-9 \f(1)=1^2-9=-8 \f(3)=3^2-9=9-9=0$

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

Question3

Graph the function and identify the roots.

Question12

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

Question2

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

Question3

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

Question6

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

Question5

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

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,

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 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

Graph the function and identify its roots.

Question5

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

Question2

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

Question3

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

Question4

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

Screen shot 2016 01 13 at 9.50.10 am

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

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

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

since

$b=25\Rightarrow \sqrt{b}=\pm5$

thus the simplified, factored form is,

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.

$\x-5=0 \x=5$ $\x+5=0 \x=-5$

Step 3: Create a table of pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \ \hline -5& f(-5) & 0 \ -2 & f(-2) & -21 \ 0&f(0)& -25\ 2&f(2)&-21\5&f(5)&0 \\hline \end{tabular}$

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

$\f(-5)=(-5)^2-25=25-25=0 \f(-2)=(-2)^2-25=4-25=-21 \f(0)=0^2-25=-25 \f(2)=2^2-25=4-25=-21 \f(5)=5^2-25=25-25=0$

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

Question5

Graph the function and identify its roots.

Question6

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

Question4

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

Question3

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

Question5

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

Screen shot 2016 01 13 at 12.16.31 pm

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

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,

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 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

Graph the function and identify the roots.

Question9

$\text{Roots: }x=\pm\sqrt{2}$

Question8

$\text{Roots: }x=\pm\sqrt{2}$

Question7

$\text{Roots: }x=\pm1$

Question3

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

Question4

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

Explanation

his 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 techniques to solve for the roots.

$\2x^2-4=0 \2x^2=4 \x^2=2 \x=\pm\sqrt{2}$

Step 2: Create a table of pairs.

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

Step 3: The values in the table are found by substituting in the x values into the function as follows.

$\f(-2)=2(-2)^2-4=2(4)-4=8-4=4 \f(-1)=2(-1)^2-4=2(1)-4=2-4=-2 \f(0)=2(0)^2-4=0-4=-4$

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

Question9

Graph the function and identify its roots.

Question4

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

Question3

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

Screen shot 2016 01 13 at 12.16.31 pm

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

Question2

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

Screen shot 2016 01 13 at 12.16.52 pm

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

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,

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 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

Graph the function and identify the roots.

Question8

$\text{Roots: }x=\pm1$

Question5

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

Question6

$\text{Roots: }x=\pm1$

Question7

$\text{Roots: }x=\pm\frac{1}{5}$

Question4

$\text{Roots: }x=\pm1$

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

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

since

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

thus the simplified, factored form is,

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.

$\x-1=0 \x=1$ $\x+1=0 \x=-1$

Step 3: Create a table of pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \ \hline -2& f(-2) & 3 \ -1 & f(-1) & 0 \ 0&f(0)& -1\ 1&f(1)&0\5&f(2)&3 \\hline \end{tabular}$

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

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

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

Question8

Graph the function and identify the roots.

Question7

$\text{Roots: }x=\pm\frac{1}{5}$

Question6

$\text{Roots: }x=\pm\frac{5}{2}$

Question5

$\text{Roots: }x=\pm{5}$

Question2

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

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

$\a=25\Rightarrow \sqrt{a}=\pm5 \b=1\Rightarrow \sqrt{b}=\pm1$

thus the simplified, factored form is,

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.

$\5x-1=0 \5x=1\\x=\frac{1}{5}$ $\5x+1=0 \5x=-1 \\x=-\frac{1}{5}$

Step 3: Create a table of pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \ \hline -1& f(-1) & 24 \ 0 & f(0) & -1 \ 1/5&f(1/5)& 0\ 1&f(1)&24\\hline \end{tabular}$

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

$\f(-1)=25(-1)^2-1=25-1=24 \f(0)=25(0)^2-1=0-1=-1 \f(1/5)=25(1/5)^2-1=25(1/25)-1=1-1=0 \f(1)=25(1)^2-1=25-1=24$

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

Question7

Graph the function and identify its roots.

Question2

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

Screen shot 2016 01 13 at 12.16.31 pm

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

Screen shot 2016 01 13 at 9.50.10 am

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

Question4

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

Question3

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

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

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

since

$b=4\Rightarrow \sqrt{b}=\pm2$

thus the simplified, factored form is,

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.

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

Step 3: Create a table of pairs.

$\begin{tabular}{ |c|c|c|c|c|c|c| } \hline x & f(x) & y \ \hline -2& f(-2) & 0 \ -1 & f(-1) & -3 \ 0&f(0)& -4\ 1&f(1)&-3\2&f(2)&0 \ 3&f(3)&5 \\hline \end{tabular}$

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

$\f(-2)=x^2-4=(-2)^2-4=4-4=0 \f(-1)=(-1)^2-4=1-4=-3 \f(0)=0^2-4=-4 \f(1)=1^2-4=1-4=-3 \f(2)=2^2-4=4-4=0 \f(3)=3^2-4=9-4=5$

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

Question2

Page 1 of 2

Return to subject