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Common Core: High School - Functions : High School: Functions

Study concepts, example questions & explanations for Common Core: High School - Functions

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All Common Core: High School - Functions Resources

6 Diagnostic Tests 82 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Graph Linear And Quadratic Functions: Ccss.Math.Content.Hsf If.C.7a

Screen shot 2016 01 23 at 7.49.43 am

What is the -intercept of the function that is depicted in the graph above?

Possible Answers:

$\textup{y-intercept}=-1$

$\textup{y-intercept}=3$

$\textup{y-intercept}=1$

$\textup{y-intercept}=-2$

$\textup{y-intercept}=2$

Correct answer:

$\textup{y-intercept}=2$

Explanation:

This question tests one's ability to recognize algebraic characteristics of a graph. This particular question examines a quadratic function.

For the purpose of Common Core Standards, "graph linear and quadratic functions and show intercepts, maxima, and minima" 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 general algebraic function for the given graph.

Since the graph is that of a parabola opening up, the general algebraic form of the function is,

y=ax^2+bx+c

where

$\\a=\textup{The width/pitch of the parabola} \\b=\textup{Relates to the vertex}=\frac{-b}{2a} \\c=\textup{Vertical Shift}$

Recall that if is negative the parabola opens down and if is positive then the parabola opens up. Also, if 0<a<1 then the width of the parabola is wider; if a>1 then the parabola is narrower.

Step 2: Identify where the graph crosses the -axis.

Screen shot 2016 01 23 at 7.49.43 am

For the function above, the vertex is also the minimum of the function and lies at the -intercept of the graph.

c=2

Therefore the vertex lies at (0,2) which means the -intercept is two.

Step 3: Answer the question.

The -intercept is two.

Report an Error

Example Question #9 : Graph Linear And Quadratic Functions: Ccss.Math.Content.Hsf If.C.7a

Screen shot 2016 01 23 at 7.49.20 am

What is the -intercept of the function that is depicted in the graph above?

Possible Answers:

$\textup{y-intercept}=1$

$\textup{y-intercept}=2$

$\textup{y-intercept}=0$

$\textup{y-intercept}=3$

$\textup{y-intercept}=-1$

Correct answer:

$\textup{y-intercept}=0$

Explanation:

This question tests one's ability to recognize algebraic characteristics of a graph. This particular question examines a quadratic 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: Identify the general algebraic function for the given graph.

Since the graph is that of a parabola opening up, the general algebraic form of the function is,

y=ax^2+bx+c

where

$\\a=\textup{The width/pitch of the parabola} \\b=\textup{Relates to the vertex}=\frac{-b}{2a} \\c=\textup{Vertical Shift}$

Recall that if is negative the parabola opens down and if is positive then the parabola opens up. Also, if 0<a<1 then the width of the parabola is wider; if a>1 then the parabola is narrower.

Step 2: Identify where the graph crosses the -axis.

Screen shot 2016 01 23 at 7.49.20 am

For the function above, the vertex is also the minimum of the function and lies at the -intercept of the graph.

c=0

Therefore the vertex lies at (0,0) which means the -intercept is zero.

Step 3: Answer the question.

The -intercept is zero.

Report an Error

Example Question #81 : High School: Functions

Screen shot 2016 01 23 at 7.50.16 am

What is the -intercept of the function that is depicted in the graph above?

Possible Answers:

$\textup{y-intercept}=4$

$\textup{y-intercept}=2$

$\textup{y-intercept}=1$

$\textup{y-intercept}=0$

$\textup{y-intercept}=8$

Correct answer:

$\textup{y-intercept}=4$

Explanation:

This question tests one's ability to recognize algebraic characteristics of a graph. This particular question examines a quadratic 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: Identify the general algebraic function for the given graph.

Since the graph is that of a parabola opening up, the general algebraic form of the function is,

y=ax^2+bx+c

where

$\\a=\textup{The width/pitch of the parabola} \\b=\textup{Relates to the vertex}=\frac{-b}{2a} \\c=\textup{Vertical Shift}$

Recall that if is negative the parabola opens down and if is positive then the parabola opens up. Also, if 0<a<1 then the width of the parabola is wider; if a>1 then the parabola is narrower.

Step 2: Identify where the graph crosses the -axis.

Screen shot 2016 01 23 at 7.50.16 am

For the function above, the parabola is shifted to the right therefore the -intercept of the graph is not at the vertex.

Therefore the -intercept lies at the point (0,4) which means the -intercept is four.

Step 3: Answer the question.

The -intercept is four.

Report an Error

Example Question #82 : High School: Functions

Screen shot 2016 01 23 at 7.50.54 am

What is the -intercept of the function that is depicted in the graph above?

Possible Answers:

$\textup{y-intercept}=-1$

$\textup{y-intercept}=2$

$\textup{y-intercept}=0$

$\textup{y-intercept}=4$

$\textup{y-intercept}=1$

Correct answer:

$\textup{y-intercept}=1$

Explanation:

This question tests one's ability to recognize algebraic characteristics of a graph. This particular question examines a quadratic 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: Identify the general algebraic function for the given graph.

Since the graph is that of a parabola opening up, the general algebraic form of the function is,

y=ax^2+bx+c

where

$\\a=\textup{The width/pitch of the parabola} \\b=\textup{Relates to the vertex}=\frac{-b}{2a} \\c=\textup{Vertical Shift}$

Recall that if is negative the parabola opens down and if is positive then the parabola opens up. Also, if 0<a<1 then the width of the parabola is wider; if a>1 then the parabola is narrower.

Step 2: Identify where the graph crosses the -axis.

Screen shot 2016 01 23 at 7.50.54 am

For the function above, the parabola is shifted to the right therefore the -intercept of the graph is not at the vertex.

Therefore the -intercept lies at the point (0,1) which means the -intercept is one.

Step 3: Answer the question.

The -intercept is one.

Report an Error

Example Question #1 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Graph the following function

$f(x)=\sqrt{x+4}$

Possible Answers:

Screen shot 2016 01 13 at 9.26.03 am

Screen shot 2016 01 13 at 9.26.38 am

Screen shot 2016 01 13 at 9.26.23 am

Screen shot 2016 01 12 at 3.00.23 pm

Screen shot 2016 01 13 at 9.25.36 am

Correct answer:

Screen shot 2016 01 12 at 3.00.23 pm

Explanation:

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

For the purpose of Common Core Standards, "graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions" 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: Make a table of (x,y) coordinates for the function.

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

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

$\\f(-4)=\sqrt{-4+4}=\sqrt{0}=0 \\f(0)=\sqrt{0+4}=\sqrt{4}=2 \\f(5)=\sqrt{5+4}=\sqrt{9}=3$

Step 2: Plot the points on a coordinate grid

Screen shot 2016 01 13 at 9.20.36 am

Step 3: Connect the points with a smooth curve.

Recall that a square root function cannot have negative values under the radical therefore, no x values less than negative four will be in the domain.

Screen shot 2016 01 12 at 3.00.23 pm

Report an Error

Example Question #1 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Graph the following function.

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

Possible Answers:

Screen shot 2016 01 13 at 9.26.38 am

Screen shot 2016 01 13 at 9.25.36 am

Screen shot 2016 01 12 at 3.00.23 pm

Screen shot 2016 01 23 at 8.33.58 am

Screen shot 2016 01 13 at 9.26.03 am

Correct answer:

Screen shot 2016 01 23 at 8.33.58 am

Explanation:

This question tests one's ability to graph a square root 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: Make a table of (x,y) coordinates for the function.

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

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

$\\f(2)=\sqrt{-2+2}=\sqrt{0}=0 \\f(-2)=\sqrt{--2+2}=\sqrt{4}=2 \\f(-7)=\sqrt{--7+2}=\sqrt{9}=3$

Step 2: Plot the points on a coordinate grid

Screen shot 2016 01 23 at 8.48.27 am

Step 3: Connect the points with a smooth curve.

Recall that a square root function cannot have negative values under the radical therefore, no x values more than two will be in the domain.

Screen shot 2016 01 23 at 8.33.58 am

Report an Error

Example Question #2 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Graph the following function.

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

Possible Answers:

Screen shot 2016 01 12 at 3.00.23 pm

Screen shot 2016 01 23 at 8.35.09 am

Screen shot 2016 01 13 at 9.26.38 am

Screen shot 2016 01 13 at 9.25.36 am

Screen shot 2016 01 23 at 8.33.58 am

Correct answer:

Screen shot 2016 01 23 at 8.35.09 am

Explanation:

This question tests one's ability to graph a square root 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: Make a table of (x,y) coordinates for the function.

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

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

$\\f(1)=\sqrt{-2(1)+2}=\sqrt{0}=0 \\f(-1)=\sqrt{-2(-1)+2}=\sqrt{4}=2 \\f(-7)=\sqrt{-2(-7)+2}=\sqrt{16}=4$

Step 2: Plot the points on a coordinate grid

Screen shot 2016 01 25 at 4.54.36 pm

Step 3: Connect the points with a smooth curve.

Recall that a square root function cannot have negative values under the radical therefore, no x values greater than one will be in the domain.

Screen shot 2016 01 23 at 8.35.09 am

Report an Error

Example Question #3 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Graph the following function.

$f(x)=\sqrt{2x}$

Possible Answers:

Screen shot 2016 01 23 at 8.35.41 am

Screen shot 2016 01 12 at 3.00.23 pm

Screen shot 2016 01 23 at 8.35.09 am

Screen shot 2016 01 23 at 8.33.58 am

Screen shot 2016 01 13 at 9.25.36 am

Correct answer:

Screen shot 2016 01 23 at 8.35.41 am

Explanation:

This question tests one's ability to graph a square root 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: Make a table of (x,y) coordinates for the function.

$\begin{tabular}{ |c|c|c|c| } \hline x & f(x) & y \\ \hline 0& f(0) & 0 \\ 2 & f(2) & 2 \\ 8&f(8)& 4\\ \hline \end{tabular}$

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

$\\f(0)=\sqrt{2(0)}=\sqrt{0}=0 \\f(2)=\sqrt{2(2)}=\sqrt{4}=2 \\f(8)=\sqrt{2(8)}=\sqrt{16}=4$

Step 2: Plot the points on a coordinate grid and connect them with a smooth curve.

Recall that a square root function cannot have negative values under the radical therefore, no x values less than zero will be in the domain.

Screen shot 2016 01 23 at 8.35.41 am

Report an Error

Example Question #4 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Graph the following function.

$f(x)=\sqrt{x+2}$

Possible Answers:

Screen shot 2016 01 12 at 3.00.23 pm

Screen shot 2016 01 23 at 8.35.41 am

Screen shot 2016 01 13 at 9.26.38 am

Screen shot 2016 01 23 at 8.33.58 am

Screen shot 2016 01 23 at 8.36.02 am

Correct answer:

Screen shot 2016 01 23 at 8.36.02 am

Explanation:

This question tests one's ability to graph a square root 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: Make a table of (x,y) coordinates for the function.

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

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

$\\f(-2)=\sqrt{2+-2}=\sqrt{0}=0 \\f(2)=\sqrt{2+2}=\sqrt{4}=2 \\f(7)=\sqrt{7+2}=\sqrt{9}=3$

Step 2: Plot the points on a coordinate grid and connect them with a smooth curve.

Recall that a square root function cannot have negative values under the radical therefore, no x values less than negative two will be in the domain.

Screen shot 2016 01 23 at 8.36.02 am

Report an Error

Example Question #5 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Graph the following function.

$f(x)=\sqrt{x-2}$

Possible Answers:

Screen shot 2016 01 23 at 8.35.09 am

Screen shot 2016 01 12 at 3.00.23 pm

Screen shot 2016 01 23 at 8.36.21 am

Screen shot 2016 01 13 at 9.25.36 am

Screen shot 2016 01 13 at 9.26.38 am

Correct answer:

Screen shot 2016 01 23 at 8.36.21 am

Explanation:

This question tests one's ability to graph a square root 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: Make a table of (x,y) coordinates for the function.

$\begin{tabular}{ |c|c|c|c| } \hline x & f(x) & y \\ \hline 2& f(2) & 0 \\ 6 & f(6) & 2 \\ 18&f(18)& 4\\ \hline \end{tabular}$

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

$\\f(2)=\sqrt{2-2}=\sqrt{0}=0 \\f(6)=\sqrt{6-2}=\sqrt{4}=2 \\f(18)=\sqrt{18-2}=\sqrt{16}=4$

Step 2: Plot the points on a coordinate grid and connect them with a smooth curve.

Recall that a square root function cannot have negative values under the radical therefore, no x values less than two will be in the domain.

Screen shot 2016 01 23 at 8.36.21 am

Report an Error

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All Common Core: High School - Functions Resources

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Common Core: High School - Functions : High School: Functions

Study concepts, example questions & explanations for Common Core: High School - Functions

All Common Core: High School - Functions Resources

Example Questions

Example Question #1 : Graph Linear And Quadratic Functions: Ccss.Math.Content.Hsf If.C.7a

Example Question #9 : Graph Linear And Quadratic Functions: Ccss.Math.Content.Hsf If.C.7a

Example Question #81 : High School: Functions

Example Question #82 : High School: Functions

Example Question #1 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Example Question #1 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Example Question #2 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Example Question #3 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Example Question #4 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

Example Question #5 : Graph Square Root, Cube Root, And Piecewise Functions: Ccss.Math.Content.Hsf If.C.7b

All Common Core: High School - Functions Resources

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