Graph Exponential, Logarithmic, and Trigonometric Functions: CCSS.Math.Content.HSF-IF.C.7e - Common Core: High School - Functions

Card 0 of 12

Question

Graph the following function.

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify the -intercept.

Recall that is just a constant,

$e\approx2.718$

and any constant raised to zero equals one.

$\f(x)=e^x \f(0)=e^0 \f(0)=1$

Therefore the -intercept occurs at the point

Step 2: Graph the function using technology and plot the -intercept.

Compare your answer with the correct one above

Question

Graph the function.

$f(x)=\ln(x)$

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify any intercepts.

Recall that the natural log cannot be less than zero and it has a x-intercept of one.

$\ln(1)=0$

Also recall that the natural log and cancel each other out which results in,

$\\ln(e)=1 \\ln(2.718)=1$

Step 2: Graph the function using technology and plot the intercept.

Compare your answer with the correct one above

Question

Graph the function.

$f(x)=2\ln(x)$

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify any intercepts.

Recall that the natural log cannot be less than zero and it has a x-intercept of one.

$\ln(1)=0$

Also recall that the natural log and cancel each other out which results in,

$\\ln(e)=1 \\ln(2.718)=1$

Since the function is,

$f(x)=2\ln(x)$

a table can be created to find the function values.

$\begin{tabular}{ c c } x & f(x) \ \hline 1& 0 \ 2& 1.386\ 4&2.773\ 5&3.219 \end{tabular}$

Step 2: Graph the function using technology and plot the intercept.

Compare your answer with the correct one above

Question

Graph the function.

$f(x)=\ln(x-1)$

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify any intercepts.

Recall that the natural log cannot be less than zero and it has a x-intercept of one.

$\ln(1)=0$

therefore,

$\ln(x-1)=0 \rightarrow x=2$

Since the function is,

$f(x)=\ln(x-1)$

a table can be created to find the function values.

$\begin{tabular}{ c c } x & f(x) \ \hline 2& 0 \ 4& 1.0986\ 6&1.609\ 2&2.079 \end{tabular}$

Step 2: Graph the function using technology and plot the intercept.

Compare your answer with the correct one above

Question

Graph the function.

$f(x)=\ln(x+1)$

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify any intercepts.

Recall that the natural log cannot be less than zero and it has a x-intercept of one.

$\ln(1)=0$

therefore,

$\ln(x+1)=0 \rightarrow x=0$

Since the function is,

$f(x)=\ln(x+1)$

a table can be created to find the function values.

$\begin{tabular}{ c c } x & f(x) \ \hline 0& 0 \ 2& 1.0986\ 4&1.609\ 6&1.946 \end{tabular}$

Step 2: Graph the function using technology and plot the intercept.

Compare your answer with the correct one above

Question

Graph the function.

$f(x)=\ln(2x)$

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify any intercepts.

Recall that the natural log cannot be less than zero and it has a x-intercept of one.

$\ln(1)=0$

therefore,

$\ln(2x)=0 \rightarrow x=\frac{1}{2}$

Since the function is,

$f(x)=\ln(2x)$

a table can be created to find the function values.

$\begin{tabular}{ c c } x & f(x) \ \hline 1/2& 0 \ 1& 0.693\ 2&1.386\ 3&1.792 \end{tabular}$

Step 2: Graph the function using technology and plot the intercept.

Compare your answer with the correct one above

Question

Graph the function.

$f(x)=\ln(x-3)$

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify any intercepts.

Recall that the natural log cannot be less than zero and it has a x-intercept of one.

$\ln(1)=0$

therefore,

$\ln(x-3)=0 \rightarrow x=4$

Since the function is,

$f(x)=\ln(x-3)$

a table can be created to find the function values.

$\begin{tabular}{ c c } x & f(x) \ \hline 3.5& -0.693 \ 4& 0\ 6&1.099\ 7&1.386 \end{tabular}$

Step 2: Graph the function using technology and plot the intercept.

Compare your answer with the correct one above

Question

Graph the function.

$f(x)=e^{2x}$

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify the -intercept.

Recall that is just a constant,

$e\approx2.718$

and any constant raised to zero equals one.

$\f(x)=e^{2x} \f(0)=e^{2\cdot0} \f(0)=e^0\f(0)=1$

Therefore the -intercept occurs at the point

Step 2: Graph the function using technology and plot the -intercept.

$\begin{tabular}{ c c } x & y \ \hline 0& 1 \ 1& 7.389\ 2&54.598 \end{tabular}$

Compare your answer with the correct one above

Question

Graph the function.

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify the -intercept.

Recall that is just a constant,

$e\approx2.718$

and any constant raised to zero equals one.

$\f(x)=e^{x} \f(0)=e^0\f(0)=1$

Since the function is

Therefore the -intercept occurs at the point

Note that this vertical shift moves the horizontal asymptote up one from zero to one.

Step 2: Graph the function using technology and plot the -intercept.

$\begin{tabular}{ c c } x & y \ \hline -1& 1.368 \ 0& 2\ 3&21.0855 \end{tabular}$

Compare your answer with the correct one above

Question

Graph the function.

$f(x)=e^{x+1}$

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify the -intercept.

Recall that is just a constant,

$e\approx2.718$

and any constant raised to zero equals one.

$\f(x)=e^{x+1} \rightarrow x=-1 \f(-1)=e^{-1+1} \f(-1)=e^0 \f(-1)=1$

Since the function is

$\f(x)=e^{x+1} \f(0)=e^0+1 \f(0)=e \f(0)=2.718$

Therefore the -intercept occurs at the point

Step 2: Graph the function using technology and plot the -intercept.

$\begin{tabular}{ c c } x & y \ \hline -1& 1 \ 0& 2.718\ 1&7.389 \end{tabular}$

Compare your answer with the correct one above

Question

Graph the function.

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify the -intercept.

Recall that is just a constant,

$e\approx2.718$

and any constant raised to zero equals one.

$\f(x)=e^{x} \f(0)=e^0\f(0)=1$

Since the function is

Therefore the -intercept occurs at the point

Note that this vertical shift moves the horizontal asymptote up two from zero to two.

Step 2: Graph the function using technology and plot the -intercept.

$\begin{tabular}{ c c } x & y \ \hline -1& 2.369 \ 0& 3\ 1&4.718 \end{tabular}$

Compare your answer with the correct one above

Question

Graph the function.

Answer

This question is testing one's ability to graph exponential functions.

For the purpose of Common Core Standards, "graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude." 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: Identify the -intercept.

Recall that is just a constant,

$e\approx2.718$

and any constant raised to zero equals one.

$\f(x)=e^{x} \ \f(0)=e^0\f(0)=1$

Since the function is

Therefore the -intercept occurs at the point

Note that this vertical shift moves the horizontal asymptote up one from zero to one.

Step 2: Graph the function using technology and plot the -intercept.

$\begin{tabular}{ c c } x & y \ \hline -1& -0.632 \ 0& 0\ 2&6.389 \end{tabular}$

Compare your answer with the correct one above

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