HSF.BF.B.3: Transformations of Graphs and Even and Odd Functions
In plain English: HSF.BF.B.3 is the Common Core functions standard that asks students to identify how replacing f(x) by f(x) + k, k f(x), f(kx) or f(x + k) changes a graph, for positive and negative k, and to find k from two graphs. Students explore the effects with graphing technology and recognize even and odd functions from graphs and formulas. It is usually taught across Algebra I and Algebra II.
Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them.
Common Core State Standards for Mathematics · Domain: Building Functions (BF) · Cluster: Build new functions from existing functions Also written as HSF-BF.B.3 or F-BF.3 · Official standard
Students study four ways to build a new function from a function f: adding k to the output, f(x) + k; multiplying the output by k, k f(x); multiplying the input by k, f(kx); and adding k to the input, f(x + k). For each, they use graphing technology with a slider for k to see what happens for positive and negative values, then explain the effect by tracking where individual points go. Changes to the output act vertically and match intuition; changes to the input act horizontally and feel backwards, which is where the lesson spends the most time.
Students then reverse the process and find k when they are given the graph of f and the transformed graph. The lesson ends with even and odd functions: an even function's graph is symmetric about the y-axis, so f(-x) = f(x), and an odd function's graph is symmetric about the origin, so f(-x) = -f(x). Students recognize both from graphs and prove them from algebraic expressions.
Learning Objectives
By the end of this lesson, students will be able to:
Describe the effect on the graph of f(x) + k, k f(x), f(kx) and f(x + k) for specific positive and negative values of k
Find the value of k when given the graph of f and the graph of the transformed function
Use graphing technology to experiment with k and explain each effect by tracking where points move
Recognize even and odd functions from their graphs and from their algebraic expressions
Prior Knowledge Required
Students should already be comfortable with:
Function notation and evaluating f at an expression such as f(x + 2) HSF.IF.A.2
Graphs of linear, quadratic, absolute value, square root and cube functions HSF.IF.C.7
Translations and reflections of figures in the coordinate plane 8.G.A.3
Give each student a table with x = -3, -2, -1, 0, 1, 2 and three columns: x², x² + 3 and (x + 3)².
Warm-Up Prompt
"Fill in the table and sketch all three graphs on one set of axes. Before you graph, predict: which way does adding 3 move the parabola in each case? After graphing, check your predictions. Were you surprised by one of them?"
Take a quick vote before graphing. Many students predict that (x + 3)² moves the parabola up or to the right. The table shows otherwise: x² + 3 has vertex (0, 3), so the graph moves up 3, while (x + 3)² is 0 at x = -3, so its vertex is (-3, 0) and the graph moves left 3. Ask: "For (x + 3)² to equal 4, what must x be?" (x = -1 or x = -5, each 3 less than the inputs 2 and -2 that give 4 for x².) This is the idea behind every horizontal transformation.
Direct Instruction25 minutes
Part 1: The four transformations. Project a graphing tool with f and a slider for k, and run through the cases in this order. For each one, track one point (a, b) on the graph of f:
f(x) + k: every output increases by k. The point (a, b) moves to (a, b + k): up if k > 0, down if k < 0.
k f(x): every output is multiplied by k. The point (a, b) moves to (a, kb). If |k| > 1 the graph stretches away from the x-axis; if 0 < |k| < 1 it compresses toward it; if k < 0 it also reflects across the x-axis.
f(x + k): the new function reaches the old output b when x + k = a, that is at x = a - k. The point (a, b) moves to (a - k, b): left if k > 0, right if k < 0.
f(kx): the new function reaches b when kx = a, that is at x = a/k. The point (a, b) moves to (a/k, b). If |k| > 1 the graph compresses toward the y-axis; if 0 < |k| < 1 it stretches away from it; if k < 0 it also reflects across the y-axis.
Check any description by substituting: if (a/k, b) should be on the graph of f(kx), then f(k · a/k) = f(a) = b.
Work through the examples below with Diagram 1. For each, name the case, the value of k and its sign before describing the graph.
f(x) + k with k negative
Compare g(x) = |x| - 3 with f(x) = |x|.
Equation: k = -3: the graph moves down 3, vertex (0, 0) to (0, -3)
f(x + k) with k positive
Compare g(x) = (x + 4)² with f(x) = x².
Equation: k = 4: the graph moves left 4, vertex (0, 0) to (-4, 0)
k f(x) with k negative
Compare g(x) = -2√x with f(x) = √x, tracking the point (4, 2).
Equation: k = -2: stretch by 2 and reflect across the x-axis, (4, 2) to (4, -4)
f(kx) with k greater than 1
For f(x) = x² - 4x, compare g(x) = f(2x) = 4x² - 8x with f, tracking the zeros and the vertex.
Equation: k = 2: compress toward the y-axis by 1/2, zeros 0 and 4 to 0 and 2, vertex (2, -4) to (1, -4)
Finding k from the graphs
The graph of g is the graph of f(x) = |x| moved so that its vertex is at (5, 0), and g(x) = f(x + k). Find k.
Equation: g(x) = |x + k| has vertex at x = -k, so -k = 5 and k = -5
Part 2: Even and odd functions. Use Diagram 2. Replacing x by -x is the case f(kx) with k = -1, a reflection across the y-axis. If that reflection leaves the graph unchanged, f(-x) = f(x) for every x and the function is even. If reflecting across the y-axis gives the same graph as reflecting across the x-axis, then f(-x) = -f(x) and the function is odd; its graph is symmetric about the origin. Show the algebraic test on f(x) = x⁴ - 5x²: f(-x) = (-x)⁴ - 5(-x)² = x⁴ - 5x², so f is even. Stress that most functions are neither, and that one pair of points is enough to show a function is not even or not odd, but a single pair can never prove that it is.
Guided Practice15-20 minutes
Pairs use a graphing tool to test each prediction after they write it down:
Describe g(x) = √(x - 5) + 1 compared with f(x) = √x. (Right 5 and up 1; the starting point (0, 0) moves to (5, 1).)
Describe g(x) = 0.5|x| compared with f(x) = |x|. (Vertical compression by 1/2; (4, 4) moves to (4, 2).)
Graph g(x) = (-x)³ and compare it with x³. (A reflection across the y-axis; it is the same graph as -x³ because x³ is odd.)
The graph of g(x) = k · x² passes through (2, 12). Find k. (4k = 12, so k = 3.)
Listen for these errors: moving f(x - 5) to the left, describing k f(x) and f(kx) with the same words, and calling a function odd because its formula has an odd power in it.
Independent Practice15 minutes
Students work alone, then check with a graphing tool:
Describe g(x) = |x + 1| - 2 compared with f(x) = |x| and give its vertex. (Left 1 and down 2; (-1, -2).)
Describe g(x) = -3x² compared with f(x) = x² and give the image of (1, 1). (Stretch by 3 and reflect across the x-axis; (1, -3).)
Describe g(x) = √(x/4) compared with f(x) = √x and give the image of (1, 1). (f(kx) with k = 1/4: horizontal stretch by 4; (4, 1).)
Decide whether each is even, odd or neither: p(x) = 4x - x³ and q(x) = 3x² + |x|. (p is odd; q is even.)
The graph of g(x) = x³ + k passes through (2, 5). Find k. (8 + k = 5, so k = -3.)
Closure5-10 minutes
Exit ticket: (1) How does the graph of f(x - 6) compare with the graph of f? (Right 6.) (2) The graph of g(x) = x² + k has vertex (0, 7). Find k. (k = 7.) (3) Is h(x) = x² + 5x even? Give a reason. (No: h(1) = 6 but h(-1) = -4.)
Differentiation Strategies
For Struggling Students
Give a four-row reference card with one tracked point for each case, such as (a, b) to (a - k, b) for f(x + k)
Have students make a table of values for f and the transformed function side by side before graphing
For horizontal cases, ask "What input gives the same output as before?" instead of relying on a rule
For Advanced Students
Show that for f(x) = x², the graphs of f(2x) and 4f(x) are identical, and explain why that does not happen for f(x) = x³ + 1
Prove that the product of two odd functions is even and the sum of two odd functions is odd
Find a function that is both even and odd, and explain why it is the only one
Assessment Guidance
What to Look For
Check that students track a specific point rather than reciting rules: a student who can say that (4, 2) moves to (2, 2) under f(2x) understands the horizontal compression. Look for correct direction words for negative k in every case. When finding k, students should substitute a point from the transformed graph and check the answer with a second point. For even and odd functions, check that students compute f(-x) for the whole expression and compare it with both f(x) and -f(x).
02
Classroom Activities
3 Activities
1
Slider Lab
25 minPairs
Pairs use graphing technology with a slider for k to experiment with all four cases on f(x) = √x, record where the point (4, 2) moves, and write an explanation of each effect. This is the technology part of the standard: students test their predictions and then explain them in terms of inputs and outputs.
Setup
Enter f(x) = √x, a slider k from -3 to 3, and one of the four transformed functions at a time
Mark the point (4, 2) on f and add the point it moves to, so students can watch it as k changes
Recording Table (with answers)
f(x) + k: k = 2 gives (4, 4); k = -2 gives (4, 0)
k f(x): k = 2 gives (4, 4); k = -2 gives (4, -4); k = 0.5 gives (4, 1)
f(kx): k = 2 gives (2, 2); k = -2 gives (-2, 2); k = 0.5 gives (8, 2)
f(x + k): k = 2 gives (2, 2); k = -2 gives (6, 2)
Discussion Questions
For k = 2, f(x) + k and k f(x) both send (4, 2) to (4, 4). Do they give the same graph? Test the point (1, 1).
What happens to the graph of f(kx) when k = 0? Why?
Write one sentence that explains why f(x + 2) moves the graph left, using the words input and output.
Modification for Distance Learning
Share a prepared Desmos graph with the slider already set up. Pairs record their table in a shared document and post one screenshot per case.
2
Find k: Graph Card Match
20 minGroups of 3-4
Each group gets 8 graph cards. Every card shows the parent graph (dashed), a transformed graph (solid) with the key points listed below, and the form of the transformation. Groups find k for each card and check it by substituting a second point.
Card 3: f(x) = |x|, g(x) = k f(x), passes through (2, 8). k = 4
Card 4: f(x) = |x|, g(x) = k f(x), passes through (-3, -1). k = -1/3
Card 5: f(x) = √x, g(x) = f(kx), passes through (1, 2). k = 4
Card 6: f(x) = √x, g(x) = f(kx), starts at (0, 0) and passes through (-9, 3). k = -1
Card 7: f(x) = x³, g(x) = f(x + k), center point (-2, 0). k = 2
Card 8: f(x) = x³, g(x) = f(x) + k, passes through (1, 4). k = 3
Procedure
Each student solves two cards, then passes them to the left for a check with a second point
Groups sort the cards by case and by the sign of k
Groups explain Card 2 to the class: why is k negative when the graph moved right?
Challenge Variation
Groups make two new cards for each other in which k is a fraction, and include one card where two different values of k give the same graph.
3
Even, Odd or Neither Sort
15 minPairs
Pairs sort 12 cards, 7 algebraic expressions and 5 graph descriptions, into three piles: even, odd and neither. For each expression they compute f(-x); for each graph they test one pair of mirror points.
Cards (with answers)
Expressions: x⁴ - 3x² (even); 2x³ + x (odd); x² + x (neither); |x| + 1 (even); x⁵ - x³ (odd); (x - 1)² (neither); the constant -7 (even)
Graphs: a parabola with vertex (0, -2) opening up (even); the line y = -2x (odd); the line y = x + 3 (neither); the graph of y = x³ moved up 1 (neither); a curve through (1, 3), (-1, -3), (2, -1) and (-2, 1), symmetric about the origin (odd)
Discussion Questions
(x - 1)² has only even powers when written as a square. Why is it not even? (Expand it: x² - 2x + 1 has an odd power.)
Can the graph of an odd function miss the origin if 0 is in its domain? (No: f(0) = -f(0) forces f(0) = 0.)
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Four Transformations
Each panel shows a parent function (dashed) and one transformation (solid), drawn to scale with grid lines 1 unit apart. The dots show where one or two points move: (0, 0) to (0, -3); (0, 0) to (-4, 0); (4, 2) to (4, -4); and (2, -4) to (1, -4) with the zero 4 moving to 2. Output changes (f(x) + k, k f(x)) act vertically; input changes (f(x + k), f(kx)) act horizontally and in the opposite direction to the sign of k.
Diagram 2: Even and Odd Functions
Left: f(x) = x² - 2 is even; (2, 2) and (-2, 2) are both on the graph, mirror images across the y-axis. Right: f(x) = x³ - 3x is odd; (2, 2) and (-2, -2) are both on the graph, a half-turn apart about the origin. Both drawn to scale for -3 ≤ x ≤ 3.
04
Homework Assignment
~30 min
HSF.BF.B.3 Homework: Transforming Graphs
Directions: For each transformation, name the case (f(x) + k, k f(x), f(kx) or f(x + k)), the value of k, and describe the effect on the graph in words. Track at least one point. Use graphing technology where a problem asks for it, and record what you entered.
Part 1: Describing and Exploring Effects (Problems 1-3)
Let f(x) = x². Describe the graph of each function compared with f and give its vertex: (a) f(x) - 5 (b) f(x - 1) (c) f(x + 6) (d) f(x) + 2.5
The graph of a function f passes through (-2, 0), (0, 4) and (3, 1). Give the images of these three points on the graph of (a) 3f(x) (b) -f(x) (c) f(2x) (d) f(-x) (e) f(x/3). Describe each effect in words.
Use a graphing calculator or Desmos. Graph f(x) = x³ - 4x, then g(x) = f(x) + k and h(x) = f(x + k) for k = -3, 0 and 2. Record where the point (2, 0) moves in each graph, and write two sentences explaining the effects you saw.
Part 2: Finding k (Problems 4-5)
(a) The graph of g(x) = f(x + k), where f(x) = √x, starts at (-7, 0). Find k. (b) The graph of g(x) = k · x² passes through (-2, -10). Find k. (c) The graph of g(x) = f(kx), where f(x) = x² - 9, has x-intercepts -1.5 and 1.5. Find every possible value of k.
The graph of f is a "tent" through (-3, 0), (0, 6) and (3, 0). (a) The graph of g(x) = f(kx) is a tent through (-1, 0), (0, 6) and (1, 0). Find k. Is there more than one possible value? (b) The graph of h(x) = k f(x) is an upside-down tent through (-3, 0), (0, -3) and (3, 0). Find k.
Part 3: Even and Odd Functions (Problem 6)
Decide algebraically whether each function is even, odd or neither, and describe the symmetry of its graph: (a) f(x) = 3x⁴ - x² + 2 (b) g(x) = x³ - 5x (c) h(x) = 2x⁵ - 3 (d) p(x) = |x| - x²
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Describing Effects
Case, k and direction all correct, with tracked points
Correct case, wrong direction or size for one part
Missing or mostly incorrect
Finding k
Correct k found and checked with a second point
Correct method, arithmetic error or a missing value
No valid method
Technology Exploration
Graphs recorded and effects explained in words
Graphs recorded without explanation
Not attempted
Even and Odd
f(-x) computed and compared with f(x) and -f(x)
Correct label without algebra
Incorrect labels
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again. A graphing tool is helpful for checking your answers.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
How does the graph of g(x) = f(x) - 7 compare with the graph of f?
Answer: B
Subtracting 7 from every output lowers every point by 7: (a, b) moves to (a, b - 7). Choices C and D describe changes to the input, such as f(x + 7) or f(x - 7).
Question 2 of 20 · Multiple Choice
How does the graph of g(x) = f(x + 5) compare with the graph of f?
Answer: A
g reaches the output f(a) when x + 5 = a, at x = a - 5, so every point moves 5 units left. Choice B is a common error: the plus sign suggests moving in the positive direction, but the input change works the opposite way.
Question 3 of 20 · Multiple Choice
What is the vertex of the graph of g(x) = |x - 8|?
Answer: C
This is f(x + k) with f(x) = |x| and k = -8, so the graph moves right 8, and the vertex (0, 0) moves to (8, 0). Check: g(8) = 0. Choice A moves the vertex the wrong way, and choices B and D move it vertically.
Question 4 of 20 · Multiple Choice
The point (2, 5) is on the graph of f. Which point must be on the graph of g(x) = 3f(x)?
Answer: D
k f(x) multiplies each output by k: g(2) = 3f(2) = 15. Choices A and B change the input, which is what f(x/3) and f(3x) would do. Choice C adds 3 instead of multiplying.
Question 5 of 20 · Multiple Choice
Which describes the graph of g(x) = -½ f(x) compared with the graph of f?
Answer: A
Multiplying outputs by -1/2 halves their size and changes their sign, so (a, b) moves to (a, -b/2): compression toward the x-axis plus a reflection across it. Choice C treats a change to the output as a change to the input. Choice D confuses multiplying by -1/2 with subtracting 1/2.
Question 6 of 20 · Multiple Choice
The point (8, -3) is on the graph of f. Which point must be on the graph of g(x) = f(4x)?
Answer: C
g(2) = f(4 · 2) = f(8) = -3, so (8, -3) moves to (8/4, -3) = (2, -3): a horizontal compression. Choice A multiplies the input by 4 instead of dividing. Choice B stretches vertically, which is 4f(x).
Question 7 of 20 · Multiple Choice
How does the graph of g(x) = f(-x) compare with the graph of f?
Answer: B
This is f(kx) with k = -1: the point (a, b) moves to (-a, b), a reflection across the y-axis. Choice A is -f(x). Choice D is true only when f is even.
Question 8 of 20 · Multiple Choice
The graph of g(x) = x² + k is the parabola y = x² moved so that its vertex is at (0, -4.5). What is k?
Answer: B
g(0) = k is the new vertex height, so k = -4.5. Choice A has the wrong sign: adding 4.5 would move the graph up.
Question 9 of 20 · Multiple Choice
The graph of g(x) = √(x + k) starts at (6, 0). What is k?
Answer: A
The graph of √x starts where the input is 0. Here x + k = 0 at x = 6, so 6 + k = 0 and k = -6. Choice B has the wrong sign: g(x) = √(x + 6) starts at (-6, 0).
Question 10 of 20 · Multiple Choice
The graph of g(x) = k · x³ passes through (2, -4). What is k?
Answer: D
g(2) = k · 8 = -4, so k = -1/2. Check with a second point: g(-2) = (-1/2)(-8) = 4. Choice A divides 8 by -4 instead of -4 by 8. Choice B confuses k with the value 2³ = 8.
Question 11 of 20 · Multiple Choice
The graph of a function is symmetric about the y-axis and passes through (3, 7). Which point must also be on the graph?
Answer: A
Symmetry about the y-axis means (a, b) and (-a, b) are both on the graph, so f is even and f(-3) = f(3) = 7. Choice B uses symmetry about the origin, which describes an odd function. Choice D reflects across y = x.
Question 12 of 20 · Multiple Choice
The graph of an odd function f passes through (-2, 5). What is f(2)?
Answer: C
For an odd function, f(-x) = -f(x), so f(2) = -f(-2) = -5. The graph is symmetric about the origin. Choice A would be correct for an even function.
Question 13 of 20 · Multiple Choice
Which function is odd?
Answer: D
f(-x) = (-x)⁵ + 2(-x) = -x⁵ - 2x = -f(x). Choice B has an odd power, but the constant 1 does not change sign: f(-1) = 0 while -f(1) = -2. Choice C is even, and choice A is neither.
Question 14 of 20 · Multiple Choice
A student graphs f(x) = x² - 1 and g(x) = f(kx) in a graphing tool with a slider for k. As k increases from 1 to 3, what happens to the x-intercepts of g?
Answer: B
The x-intercepts of g solve (kx)² = 1, so x = ±1/k. As k grows from 1 to 3, they move from ±1 to ±1/3: the graph is compressed toward the y-axis. Choice A describes f(x/k), a horizontal stretch.
Question 15 of 20 · Short Answer
Using a graphing tool, a student compares 2f(x) and f(2x) for f(x) = √x. Explain the difference between the two graphs using the point (9, 3).
For 2f(x) the outputs double, so (9, 3) moves to (9, 6): a vertical stretch by 2. For f(2x) the input 4.5 gives f(9) = 3, so (9, 3) moves to (4.5, 3): a horizontal compression by 1/2. The graphs are different: at x = 9, 2f(9) = 6 but f(18) = √18 ≈ 4.24.
Question 16 of 20 · Short Answer
Show algebraically that f(x) = 2x⁴ - 5x² is even, and describe what this means for its graph.
f(-x) = 2(-x)⁴ - 5(-x)² = 2x⁴ - 5x² = f(x) for every x, so f is even. Its graph is symmetric about the y-axis: for example, f(1) = f(-1) = -3.
Question 17 of 20 · Short Answer
Is g(x) = x³ - x² even, odd or neither? Show your reasoning.
g(-x) = -x³ - x². This equals neither g(x) = x³ - x² nor -g(x) = -x³ + x². For example, g(1) = 0 and g(-1) = -2, so g(-1) ≠ g(1) and g(-1) ≠ -g(1). g is neither even nor odd.
Question 18 of 20 · Short Answer
The graph of f(x) = |x| - 4 has x-intercepts -4 and 4. The graph of g(x) = f(kx) has x-intercepts -1 and 1. Find every possible value of k.
The intercepts of g solve |kx| = 4, so x = ±4/|k|. Setting 4/|k| = 1 gives k = 4 or k = -4. Both work because f is even: f(-4x) = f(4x), so the reflection across the y-axis does not change this graph.
Question 19 of 20 · Short Answer
Describe the graph of g(x) = (x - 3)² + 2 compared with the graph of f(x) = x², naming each case and its value of k, and give the vertex.
f(x - 3) is f(x + k) with k = -3, which moves the graph right 3. Adding 2 is f(x) + k with k = 2, which moves it up 2. The vertex moves from (0, 0) to (3, 2). Check: g(3) = 2.
Question 20 of 20 · Short Answer
A student graphs y = x³ + 2 in Desmos and says it is odd because it looks like the graph of x³. Use two points on the graph to explain whether the student is right.
(1, 3) is on the graph, so an odd function would also pass through (-1, -3). But at x = -1, y = -1 + 2 = 1, so the graph passes through (-1, 1) instead. The function is not odd (and not even, since 1 ≠ 3). Moving y = x³ up 2 destroys its symmetry about the origin: the center of symmetry is now (0, 2).
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSF.BF.B.3 mean?
HSF.BF.B.3 asks students to know what happens to a graph when f(x) is replaced by f(x) + k, k f(x), f(kx) or f(x + k), for positive and negative k, and to find k from two graphs. Students should explore the effects with graphing technology and explain them, and recognize even and odd functions from graphs and formulas.
Is HSF.BF.B.3 Algebra 1 or Algebra 2?
Both. Many course sequences introduce shifts and stretches of linear, quadratic and absolute value graphs in Algebra I, then apply the same four transformations to square root, cube, exponential and other functions in Algebra II. Even and odd functions are often introduced in Algebra II.
Why does f(x + 2) move the graph left instead of right?
Because the new function reaches each old output 2 units earlier. If f(5) = 1, then g(x) = f(x + 2) gives 1 when x + 2 = 5, that is at x = 3. Every point moves from (a, b) to (a - 2, b). Asking "which input gives the same output as before?" helps students reason about it instead of memorizing a rule.
What is the difference between k f(x) and f(kx)?
k f(x) changes the outputs, so it stretches or compresses the graph vertically: (a, b) moves to (a, kb). f(kx) changes the inputs, so it stretches or compresses the graph horizontally: (a, b) moves to (a/k, b). For k = 2, 2f(x) is twice as tall and f(2x) is half as wide.
What happens when k is negative?
For f(x) + k and f(x + k), a negative k reverses the direction of the shift: f(x) - 3 moves down and f(x - 3) moves right. For k f(x), a negative k adds a reflection across the x-axis; for f(kx), it adds a reflection across the y-axis. For example, -2f(x) is a vertical stretch by 2 followed by a flip.
How do you tell if a function is even or odd?
Compute f(-x) and simplify. If f(-x) = f(x) for every x, the function is even and its graph is symmetric about the y-axis. If f(-x) = -f(x), it is odd and its graph is symmetric about the origin. If neither is true, the function is neither. On a graph, check whether the reflection across the y-axis, or the half-turn about the origin, maps the graph onto itself.
Is a function with odd exponents always odd?
No. f(x) = x³ + 5 has an odd exponent but is not odd, because the constant term does not change sign: f(-1) = 4 but -f(1) = -6. A polynomial is odd only when every term has an odd power, and even only when every term has an even power (a constant counts as an even power, x⁰).
How is technology used for this standard?
The standard says to "experiment with cases and illustrate an explanation of the effects on the graph using technology". A graphing tool with a slider for k lets students see a graph move as k changes, including negative and fractional values. The explanation still has to come from the students: they should connect what they see to what happens to the inputs or outputs of f.
What is a common mistake when finding k from a graph?
A common one is getting the sign wrong for horizontal shifts: if the vertex moves right 3, then k = -3 in f(x + k). Another is forgetting that two values of k can give the same graph, as with f(kx) for an even f, where k and -k give identical graphs. Checking k with a second point catches both.
How does HSF.BF.B.3 connect to later topics?
The same four transformations describe the amplitude, period, phase shift and midline of trigonometric graphs (HSF.IF.C.7 and HSF.TF.B.5), and the vertex form of a quadratic. Even and odd symmetry reappears with cosine (even) and sine (odd), and in Calculus. Students who can track a point through a transformation can handle any new function type.
07
Related Standards
6 standards
These standards connect to HSF.BF.B.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.A.3Prerequisite
Describe the effect of dilations, translations, rotations and reflections using coordinates
