Graphs of Polynomial Functions

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Pre-Calculus › Graphs of Polynomial Functions

Questions 1 - 10

Give the -intercept of the graph of the function

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

Round to the nearest tenth, if applicable.

The graph has no -interceptx

Explanation

The -intercept is , where :

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

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

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

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

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

$4^{a- 3} = 0.6$

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

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

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

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

$a \approx 2.63$

The -intercept is .

Write the equation for the polynomial in the graph:

Graph 4 write funct

Explanation

The zeros of the polynomial are . That means that the factors equal zero when these values are plugged in.

The first factor is or equivalently $x - \frac{3}{5} = 0$ multiply both sides by 5:

The second and third factors are and

Multiply:

$(5x^2 -18x + 9 ) ( x + 2 ) = 5x^3 -18x^2 +9x \enspace + 10x^2 -36x +18$

Because the graph goes down-up-down instead of the standard up-down-up, the graph is negative, so change all of the signs:

Give the -intercept of the graph of the function

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

Round to the nearest tenth, if applicable.

The graph has no -interceptx

Explanation

The -intercept is , where :

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

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

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

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

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

$4^{a- 3} = 0.6$

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

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

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

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

$a \approx 2.63$

The -intercept is .

Write the equation for the polynomial in the graph:

Graph 4 write funct

Explanation

The zeros of the polynomial are . That means that the factors equal zero when these values are plugged in.

The first factor is or equivalently $x - \frac{3}{5} = 0$ multiply both sides by 5:

The second and third factors are and

Multiply:

$(5x^2 -18x + 9 ) ( x + 2 ) = 5x^3 -18x^2 +9x \enspace + 10x^2 -36x +18$

Because the graph goes down-up-down instead of the standard up-down-up, the graph is negative, so change all of the signs:

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.

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

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

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 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.

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 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 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 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

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