Pre-Calculus - Graph a Polynomial Function

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 .

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 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

Which of the following is an accurate graph of $f(x) = -(x + 4)^{2}$ ?

Varsity1

Varsity2

Varsity10

Varsity11

Varsity12

Explanation

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

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

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

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

The resulting x-axis intercept is: .

Setting tells the y-value of the y-axis intercept;

$y = -(0 + 4)^{2}$

$= -1(4)^{2}$

The resulting y-axis intercept is:

Graph the following function and identify the zeros.

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 12.17.10 pm

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

Screen shot 2016 01 13 at 12.16.52 pm

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

Screen shot 2016 01 13 at 12.16.31 pm

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

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.

Separating the function into two parts...

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

Factoring out from the first set results in...

The new factored form of the function is,

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,

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 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

Graph a Polynomial Function

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