HSF.IF.C.7: Graphing Functions and Showing Key Features
In plain English: HSF.IF.C.7 is the Common Core functions standard that asks students to graph a function from its equation and show key features: intercepts, maxima and minima, zeros, asymptotes, end behavior, and the period, midline and amplitude of trig graphs. It spans Algebra I, Algebra II and Precalculus, with hand graphs for simple cases and technology for harder ones; part d, on rational functions, is an advanced (+) standard.
Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.
a.Graph linear and quadratic functions and show intercepts, maxima, and minima.
b.Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions.
c.Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior.
d.(+) Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior.
e.Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude.
Common Core State Standards for Mathematics · Domain: Interpreting Functions (IF) · Cluster: Analyze functions using different representations Also written as HSF-IF.C.7 or F-IF.7 · Official standard
Students graph functions from their equations and show the features that describe each graph: intercepts, maxima and minima, zeros, asymptotes, end behavior, and for trigonometric graphs the period, midline and amplitude. The lesson follows the five sub-standards: linear and quadratic functions (a); square root, cube root, piecewise, step and absolute value functions (b); polynomial functions (c); rational functions, an advanced (+) topic (d); and exponential, logarithmic and trigonometric functions (e).
The habit students build is the same for every family: rewrite or factor the equation to reveal a feature, plot the feature, then sketch the rest of the graph. Simple cases are graphed by hand. When the equation does not factor nicely, students use a graphing calculator or graphing software and interpret what the screen shows. Teachers who teach this over two periods can split after part b.
Learning Objectives
By the end of this lesson, students will be able to:
Graph linear and quadratic functions and label intercepts, maxima and minima
Graph square root, cube root, piecewise, step and absolute value functions with correct endpoints and turning points
Use a factorization to find the zeros of a polynomial or rational function, and describe end behavior and asymptotes
Graph exponential and logarithmic functions with intercepts and asymptotes, and trigonometric functions with period, midline and amplitude
Decide when a graph can be drawn by hand and when technology is needed, and interpret a technology graph
Prior Knowledge Required
Students should already be comfortable with:
Function notation and evaluating functions HSF.IF.A.2
Linear functions and slope-intercept form 8.F.A.3
Factoring quadratics and polynomials HSA.SSE.A.2
Zeros of polynomials from a factorization HSA.APR.B.3
Radian measure and the unit circle, for the trigonometric part of the lesson HSF.TF.A.2
Put two equations on the board and give students three minutes with pencil and graph paper, no calculator:
Warm-Up Prompt
"Sketch y = (3/2)x - 3 and label where it crosses each axis. Then look at y = x² + 1 without graphing it: can its graph ever touch the x-axis? How do you know?"
Collect answers. The line crosses the y-axis at (0, -3) and the x-axis at (2, 0), found by setting x = 0 and then y = 0. For y = x² + 1, x² is never negative, so y is at least 1: the graph has a minimum of 1 at (0, 1) and no x-intercepts. Name the idea that drives the whole lesson: a key feature is a point or a behavior you can read from the equation, and a good graph shows it and labels it.
Direct Instruction25-30 minutes
Work through the function families in the order of the sub-standards. For each family, ask the same two questions: which features does this family have, and which form of the equation shows them fastest? Use Diagram 1 for parts a and b and Diagram 2 for parts c, d and e.
Linear and quadratic (a): intercepts from f(0) and f(x) = 0; for a quadratic, the vertex at x = -b/(2a) is a maximum when a < 0 and a minimum when a > 0.
Square root and cube root (b): y = √(x - h) + k starts at (h, k) and exists only for x ≥ h; y = ∛(x - h) + k is defined for every x and bends through its center (h, k).
Piecewise, absolute value and step (b): graph each piece only on its interval, and mark each endpoint closed (included) or open (not included). An absolute value graph is a V with its vertex at the turning point; a step graph is a row of horizontal segments.
Polynomial (c): factor to find the zeros; a factor raised to an even power touches the axis, an odd power crosses it. The leading term alone gives the end behavior.
Rational, (+) (d): factor top and bottom. Zeros come from the numerator, vertical asymptotes from the denominator, and the degrees of numerator and denominator give the end behavior.
Exponential, logarithmic and trigonometric (e): an exponential has a horizontal asymptote and a log a vertical one; find each intercept by substitution. For y = A sin(Bx) + D or y = A cos(Bx) + D, the amplitude is |A|, the period is 2π/|B| and the midline is y = D.
Quadratic function (part a)
Graph f(x) = x² - 2x - 8 and show its intercepts and its minimum.
Equation: f(x) = (x + 2)(x - 4): zeros -2 and 4, y-intercept -8, axis x = 1, minimum -9 at (1, -9)
Piecewise-defined function (part b)
Graph f(x) = x + 4 for x < -1 and f(x) = x² for x ≥ -1.
Equation: Open circle at (-1, 3), closed circle at (-1, 1); x-intercepts -4 and 0; y-intercept 0
Polynomial function (part c)
Graph f(x) = -x³ + x² + 6x using a factorization, and describe its end behavior.
Equation: f(x) = -x(x + 2)(x - 3): zeros -2, 0, 3; as x → -∞, f(x) → ∞ and as x → ∞, f(x) → -∞
Rational function, (+) (part d)
Graph r(x) = (2x - 4)/(x + 1) and show its zero, asymptotes and end behavior.
Equation: Zero x = 2, vertical asymptote x = -1, horizontal asymptote y = 2, y-intercept -4
Exponential and logarithmic functions (part e)
Graph f(x) = 2x - 4 and g(x) = log₂x + 1 on the same axes.
Equation: f: y-intercept -3, x-intercept 2, f(x) → -4 as x → -∞. g: x-intercept 1/2, no y-intercept, vertical asymptote x = 0
Finish with Diagram 2D: in y = 2 sin(πx/3) + 1, the midline is y = 1, the amplitude is 2 (so the graph runs from -1 to 3), and the period is 2π/(π/3) = 6. By hand or technology? Model one case where a factorization is not available, such as f(x) = x³ - 2x² - 5x + 3: no rational number is a zero, so graph it with technology and read the zeros as approximately -1.77, 0.52 and 3.25. The standard asks for hand graphs in simple cases and technology in harder ones; students should still say what the end behavior must be before they look at the screen.
Guided Practice20 minutes
Pairs graph four functions by hand, one at a time, and label every key feature before you confirm with a projected graph: q(x) = -(x + 1)² + 4 (vertex and maximum (-1, 4), zeros -3 and 1, y-intercept 3), s(x) = 2√x - 4 (starts at (0, -4), x-intercept 4), p(x) = (x - 1)²(x + 3) (crosses at -3, touches at 1, y-intercept 3, falls to the left and rises to the right) and the (+) function k(x) = (x + 1)/(x - 2) (zero -1, vertical asymptote x = 2, horizontal asymptote y = 1, y-intercept -1/2). Listen for these errors: reading the vertex of -(x + 1)² + 4 as (1, 4), drawing s past its starting point to the left, crossing the axis at a double zero, and joining the two branches of k across the asymptote.
Independent Practice15-20 minutes
Students graph five functions on their own and list the features next to each graph: m(x) = |2x + 4| - 2 (vertex (-2, -2), x-intercepts -3 and -1, y-intercept 2), w(x) = 3x + 1 (y-intercept 2, no x-intercept, w(x) → 1 as x → -∞), v(x) = log(x + 10) with base 10 (vertical asymptote x = -10, x-intercept -9, y-intercept 1), u(x) = cos(2x) - 3 (amplitude 1, period π, midline y = -3) and, with technology, d(x) = x⁴ - 4x² + x (zeros 0 and about -2.11, 0.25 and 1.86; rises on both ends). Students circle any feature they found with technology instead of by hand.
Closure10 minutes
Exit ticket: (1) Describe the end behavior of y = -2x⁵ + 7x. (Falls to the right and rises to the left, because the leading term -2x⁵ has odd degree and a negative coefficient.) (2) (+) Name the asymptotes and zero of y = (x - 5)/(x + 3). (Vertical x = -3, horizontal y = 1, zero 5.) (3) In one sentence, say why a step function needs open and closed circles.
Differentiation Strategies
For Struggling Students
Give a feature checklist for each family (intercepts, turning point, endpoints, asymptotes, end behavior) and have students tick off each one on their graph
Start each graph with a short table of values that includes the x-values of the key features, then connect the points
For piecewise and step functions, have students shade the interval of each piece on the x-axis before drawing it
For Advanced Students
Ask for an equation of a rational function with zeros at 1 and -2, vertical asymptote x = 3 and horizontal asymptote y = 2, then check it with technology
Ask why y = (x² - 1)/(x - 1) has a hole at x = 1 instead of an asymptote, and how to show it on the graph
Ask students to write a sine function for daylight hours in a city with 9.5 hours in December and 14.5 hours in June, and explain what the period, midline and amplitude mean
Assessment Guidance
What to Look For
A complete graph shows its features, not only a curve: labeled intercepts, a labeled maximum or minimum, open and closed endpoints on piecewise and step graphs, dashed asymptotes, arrows or words for end behavior, and a marked midline and period on trig graphs. When a student uses technology, ask them to say which features they predicted from the equation first. Watch for sign errors in shifts, such as placing the start of y = √(x + 4) - 1 at x = 4, and for graphs that cross a vertical asymptote.
02
Classroom Activities
3 Activities
1
Function Family Stations
25 minGroups of 3-4
Six stations, one per family, each with a card holding one or two functions. Groups graph each function by hand on graph paper and label its key features, then rotate every four minutes. Every station card also names the features a complete graph must show.
Station Cards (with answers for the teacher)
Station 1, linear and quadratic: y = 2x + 6 (x-intercept -3, y-intercept 6); y = x² - 6x + 8 (zeros 2 and 4, y-intercept 8, minimum -1 at (3, -1))
Station 2, roots: y = √(x - 1) + 1 (starts at (1, 1), no intercepts); y = -∛x + 2 (x-intercept 8, y-intercept 2)
Station 3, piecewise: y = 3 - x for x < 2 and y = x - 1 for x ≥ 2 (both pieces meet at (2, 1), no x-intercept, y-intercept 3). Ask: which absolute value function is this? (y = |x - 2| + 1)
Station 4, step: y = ⌊x⌋ + 2 for -2 ≤ x < 3 (five segments at heights 0 to 4, closed on the left end, open on the right end)
Station 5, polynomial: y = x²(x - 4) (touches at 0, crosses at 4, falls to the left, rises to the right)
Station 6, exponential and log: y = (1/3)x - 1 (x- and y-intercept at the origin, y → -1 as x → ∞); y = log₃x (x-intercept 1, vertical asymptote x = 0)
Procedure
At each station, one student reads the card, one plots the key features first, and one sketches the rest of the graph; rotate roles at every station
Before leaving, the group writes one sentence on a sticky note: the fastest way to find the key feature at this station
After the last rotation, groups compare their Station 3 and Station 4 graphs with another group and fix any endpoint that is open when it should be closed
Modification for Distance Learning
Post each station as a slide with a blank coordinate grid. Groups work in breakout rooms, draw on the slide, and screenshot the finished graph to a shared folder.
2
By Hand or by Technology?
20 minPairs
Pairs sort eight functions into two piles: ones they can graph accurately by hand from a factorization, and ones that need technology. They then graph one from each pile and compare what each method shows. This works on the main clause of the standard: by hand in simple cases, technology for more complicated ones.
The Eight Cards
By hand: y = (x + 2)(x - 1)(x - 3) (zeros -2, 1, 3; falls to the left, rises to the right), y = x⁴ - 5x² + 4 (zeros ±1 and ±2), y = -(x - 3)²(x + 1) (touches at 3, crosses at -1), and the (+) function y = (x² - 9)/(x² - 4) (zeros ±3, vertical asymptotes x = ±2, horizontal asymptote y = 1)
Technology: y = x³ - 2x + 5 (one real zero, about -2.09), y = 0.5x³ - 2x + 1 (zeros about -2.21, 0.54 and 1.68), y = x⁴ - 3x + 1 (two real zeros, about 0.34 and 1.31), and the (+) function y = (x + 1)/(x² - 2x - 2) (vertical asymptotes near x = -0.73 and x = 2.73)
Procedure
Pairs sort the cards and write one reason for each choice: "it factors" or "no factorization is available"
Each pair graphs one "by hand" card on paper and then checks it on a graphing calculator
Each pair graphs one "technology" card on the calculator, sketches the screen on paper, and labels the approximate zeros to two decimal places, the end behavior and any asymptotes
Discussion Questions
What can you say about y = x³ - 2x + 5 before you graph it, even without the zeros?
The calculator screen for y = (x² - 9)/(x² - 4) may show a nearly vertical line near x = 2. Is it part of the graph?
Why can a graphing window hide a key feature? Give an example.
3
Trig and Exponential Feature Match
20 minGroups of 3
Groups receive 8 equation cards, 8 graph cards and 8 feature cards and make matching triples. The feature cards list the period, midline and amplitude for trigonometric functions, and the intercepts and asymptote for exponential and log functions (part e).
Equation Cards and Matching Features
y = sin x (period 2π, midline y = 0, amplitude 1)
y = 3 cos x (period 2π, midline y = 0, amplitude 3)
y = sin(2x) + 1 (period π, midline y = 1, amplitude 1)
y = -2 sin(πx) (period 2, midline y = 0, amplitude 2)
y = cos(x/2) - 1 (period 4π, midline y = -1, amplitude 1)
y = 4 sin(πx/6) + 10 (period 12, midline y = 10, amplitude 4)
y = 5(2)x (y-intercept 5, no x-intercept, asymptote y = 0 as x → -∞)
y = log₂(x + 8) (x-intercept -7, y-intercept 3, vertical asymptote x = -8)
Procedure
Groups first match equations to feature cards using only the equation, then match the feature cards to graphs
For each trigonometric graph, a student marks the midline in color and shows one full period with a bracket
Groups explain one match to the class where two graph cards looked alike, such as y = sin x and y = -2 sin(πx)
Challenge Variation
Give groups a blank graph card and ask them to draw y = 4 sin(πx/6) + 10 as a model of temperature in degrees Celsius over a 12-hour cycle, labeling the warmest and coolest points.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Key Features for Parts a and b
Four graphs drawn to scale. A: the parabola f(x) = x² - 2x - 8 with zeros -2 and 4, y-intercept -8, axis x = 1 and minimum (1, -9). B: y = √(x + 4) - 1 starts at (-4, -1) and crosses the x-axis at -3; y = ∛(x - 2) + 1 bends through its center (2, 1) and crosses the x-axis at 1. C: h(x) = -|x - 2| + 3 has maximum (2, 3), x-intercepts -1 and 5 and y-intercept 1. D: a garage charges $4 for each hour or part of an hour, so each step has an open left end and a closed right end.
Diagram 2: Key Features for Parts c, d and e
A: f(x) = -x(x + 2)(x - 3) crosses at -2, 0 and 3; its leading term -x³ makes it rise to the left and fall to the right. B (+): r(x) = (2x - 4)/(x + 1) has zero 2, y-intercept -4, vertical asymptote x = -1 and horizontal asymptote y = 2. C: y = 2x - 4 has y-intercept -3, x-intercept 2 and asymptote y = -4; y = log₂x + 1 has x-intercept 1/2 and asymptote x = 0. D: y = 2 sin(πx/3) + 1 has midline y = 1, amplitude 2 and period 6.
04
Homework Assignment
~30 min
HSF.IF.C.7 Homework: Graphing Functions and Their Key Features
Directions: Graph each function on graph paper. Label every key feature asked for, with coordinates. Use open and closed circles for endpoints, dashed lines for asymptotes, and a sentence or arrows for end behavior. Problems 4 and 5 ask you to check your hand graph with technology.
Part 1: Parts a and b (Problems 1-3)
Graph g(x) = -(1/2)x + 3 and h(x) = x² + 2x - 15 on separate axes. Label the intercepts of both graphs and the vertex of h. Is the vertex a maximum or a minimum?
Graph f(x) = -√(x + 1) + 3 and k(x) = ∛(x + 1) - 2. For each, label the starting point or center and both intercepts.
(a) Graph p(x) = x + 3 for x < -1 and p(x) = -2x - 1 for x ≥ -1, with correct open and closed circles, and label its intercepts. (b) Graph a(x) = 2|x - 3| - 4 and label its vertex and intercepts. (c) An e-scooter costs $1.00 to unlock plus $0.40 for each minute or part of a minute. Graph the cost C(t) = 1 + 0.40⌈t⌉ for 0 < t ≤ 5 minutes and find the cost of a 3.2-minute ride.
Part 2: Parts c and d (Problems 4-5)
Factor and graph f(x) = x⁴ - 10x² + 9 and g(x) = -x³ + 4x² - 4x. Label all zeros, say where each graph crosses or touches the x-axis, and describe the end behavior of each.
(+) Graph r(x) = (x - 4)/(x² - 1). Label its zero, y-intercept, vertical asymptotes and horizontal asymptote, and describe its end behavior. Then check your graph with technology.
Part 3: Part e (Problem 6)
Graph (a) f(x) = 2x + 1 - 8 and (b) g(x) = log₂(x - 3), labeling intercepts and asymptotes and describing the end behavior. Then (c) graph h(x) = 3 sin(πx/4) + 2 for 0 ≤ x ≤ 16 and label its midline, amplitude, period, maximum and minimum.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Accurate Graphs
Shape, position and scale correct for every function
Most graphs correct, one shift or reflection error
Graphs missing or mostly incorrect
Key Features
All intercepts, extrema, zeros and asymptotes labeled with coordinates
Most features labeled, a few missing
Features not labeled
Endpoints and Behavior
Open and closed circles, end behavior and trig features all correct
One or two endpoint or end behavior errors
Endpoints and end behavior not shown
Technology Use
Hand graph checked with technology and differences explained
Technology used but not compared with the hand graph
No technology check
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Keep graph paper and a graphing calculator nearby: sketch before you choose. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again. Questions marked (+) cover the advanced part d on rational functions.
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
What are the intercepts of the graph of f(x) = 5 - (5/2)x?
Answer: B
Set x = 0: f(0) = 5, so the y-intercept is 5. Set f(x) = 0: (5/2)x = 5, so x = 2. Choice A swaps the two intercepts. Choice C comes from a sign error when solving for x, and choice D uses the slope -5/2 as if it were the y-intercept.
Question 2 of 20 · Multiple Choice
Which statement describes the graph of f(x) = -x² + 6x - 5?
Answer: A
The vertex is at x = -b/(2a) = -6/(2·(-1)) = 3, and f(3) = -9 + 18 - 5 = 4. Since a = -1 < 0, the parabola opens down, so (3, 4) is a maximum. Factoring gives f(x) = -(x - 1)(x - 5), so the zeros are 1 and 5. Choice B calls the vertex a minimum, choice C drops the negative sign in -b/(2a), and choice D reverses the signs of the zeros.
Question 3 of 20 · Short Answer
Graph f(x) = 2x² + 4x - 6 by hand. Give its zeros, y-intercept, axis of symmetry and vertex, and say whether the vertex is a maximum or a minimum.
Factor: f(x) = 2(x² + 2x - 3) = 2(x + 3)(x - 1), so the zeros are -3 and 1. The y-intercept is -6. The axis is halfway between the zeros, x = -1, and f(-1) = 2 - 4 - 6 = -8, so the vertex (-1, -8) is a minimum (a = 2 > 0).
Question 4 of 20 · Multiple Choice
Which statement is true about the graph of f(x) = √(x - 3) + 2?
Answer: C
The expression under the root must be at least 0, so x ≥ 3, and f(3) = 2: the graph starts at (3, 2). Since √(x - 3) ≥ 0, every output is at least 2, so there is no x-intercept. Choice A uses the wrong sign for the shift, choice B swaps the coordinates, and choice D comes from squaring √(x - 3) = -2, which gives x = 7, but f(7) = 4, not 0.
Question 5 of 20 · Multiple Choice
What are the intercepts of the graph of f(x) = ∛(x + 8)?
Answer: D
f(0) = ∛8 = 2, and f(x) = 0 when x + 8 = 0, so x = -8. Choice A has the wrong sign. Choice B computes √8 instead of ∛8. Choice C confuses cube roots with square roots: cube roots of negative numbers exist, such as ∛(-8) = -2, so the graph extends to the left forever.
Question 6 of 20 · Multiple Choice
Which statement describes the graph of f(x) = |x + 1| - 4?
Answer: B
The V turns where x + 1 = 0, at (-1, -4), and opens upward, so the vertex is a minimum. Solve |x + 1| = 4: x + 1 = 4 or x + 1 = -4, so x = 3 or x = -5. Choice A shifts the wrong way. Choice C calls the vertex a maximum, but the coefficient of |x + 1| is positive. Choice D uses the wrong sign for the vertical shift.
Question 7 of 20 · Multiple Choice
A library charges $0.50 for each day or part of a day that a book is late, so the fee for x days late is F(x) = 0.5⌈x⌉ for 0 < x ≤ 5. Which describes the piece of the graph over 2 < x ≤ 3?
Answer: A
Any time over 2 days and up to 3 days rounds up to 3 days, so the fee is 0.5 · 3 = $1.50, and F(3) = 1.50 is included while x = 2 is not (F(2) = 1.00). Choice B puts the open and closed ends on the wrong sides: exactly 2 days costs $1.00, not $1.50. Choice C uses the fee for the previous step, and choice D connects the steps with a slanted line, but the fee jumps instead of growing gradually.
Question 8 of 20 · Multiple Choice
For f(x) = x² when x < 1 and f(x) = 4 - x when x ≥ 1, which point is drawn as an open circle on the graph?
Answer: D
The first piece, x², is used only for x < 1, so it approaches (1, 1) without reaching it: that point is an open circle. At x = 1 the second piece applies, f(1) = 4 - 1 = 3, so (1, 3) is a closed circle (choice A). Choice B uses 4 - x at x = 0, where the first piece applies. Choice C is a real point of the graph, the x-intercept of the second piece.
Question 9 of 20 · Multiple Choice
Which describes the end behavior of f(x) = -2x⁴ + x³ + 5?
Answer: D
For large |x|, the leading term -2x⁴ decides the behavior. Its degree is even, so both ends go the same way, and its coefficient is negative, so both ends fall: as x → ±∞, f(x) → -∞. Choice A ignores the negative sign. Choices B and C describe odd-degree polynomials.
Question 10 of 20 · Multiple Choice
At which x-intercept does the graph of f(x) = x(x + 3)(x - 2)² touch the x-axis and turn around instead of crossing it?
Answer: C
The factor (x - 2)² has an even power, so f(x) does not change sign at x = 2: the graph touches the axis and turns. The factors x and (x + 3) have power 1, so the graph crosses at 0 and -3 (choices A and B). Choice D reads the zero from (x - 2)² with the wrong sign.
Question 11 of 20 · Multiple Choice
(+) Which describes the graph of r(x) = (x + 3)/(x - 2)?
Answer: B
The denominator is 0 at x = 2 and the numerator is not, so x = 2 is a vertical asymptote. The numerator is 0 at x = -3, the zero. Numerator and denominator have the same degree with leading coefficients 1 and 1, so y = 1 is the horizontal asymptote. Choice A swaps the roles of numerator and denominator. Choice C uses y = 0, which is the rule for a numerator of lower degree.
Question 12 of 20 · Multiple Choice
(+) As x → ∞, what value does r(x) = (6x - 1)/(2x + 5) approach?
Answer: A
For large x, the leading terms dominate: r(x) ≈ 6x/(2x) = 3, so the graph has horizontal asymptote y = 3. Choice B ignores the 2 in the denominator. Choice C divides the constant terms, which is r(0), the y-intercept, not the end behavior. Choice D would need a numerator of higher degree.
Question 13 of 20 · Multiple Choice
Which describes the graph of f(x) = 3 · 2x - 6?
Answer: C
f(0) = 3 · 1 - 6 = -3. Solve 3 · 2x = 6: 2x = 2, so x = 1. As x → -∞, 2x → 0, so f(x) → -6: the asymptote is y = -6. Choice A forgets that the asymptote moves down with the graph. Choice B uses 20 = 0 instead of 1, and choice D uses 20 = 2.
Question 14 of 20 · Multiple Choice
Which describes the graph of f(x) = log₃x - 2?
Answer: A
Solve log₃x = 2: x = 3² = 9. The log is defined only for x > 0, so the y-axis x = 0 is a vertical asymptote and there is no y-intercept. Choice B multiplies 3 · 2 instead of raising 3 to the power 2. Choice C gives the x-intercept of log₃x before the shift down. Choice D substitutes x = 0, but log₃0 is undefined.
Question 15 of 20 · Multiple Choice
What are the amplitude, period and midline of y = 4 sin(πx/5) - 1?
Answer: D
The amplitude is |A| = 4 and the midline is y = D = -1. The period is 2π/B = 2π/(π/5) = 10. Choice A gives the coefficient B instead of the period. Choice B mixes up the roles of the numbers. Choice C multiplies 2π by 5 but forgets to divide by π.
Question 16 of 20 · Multiple Choice
What are the maximum value and the period of y = -2 cos(3x) + 5?
Answer: B
The midline is y = 5 and the amplitude is |-2| = 2, so the graph runs from 5 - 2 = 3 to 5 + 2 = 7: the maximum is 7. The negative sign only flips the graph (it starts at its minimum). The period is 2π/3. Choice A reads the minimum as the maximum. Choice C multiplies 2π by 3 instead of dividing. Choice D gives the midline value.
Question 17 of 20 · Short Answer
Graph f(x) = -(x + 1)(x - 2)(x - 4). Give its zeros, its y-intercept and its end behavior.
Zeros: -1, 2 and 4, each a simple zero, so the graph crosses the axis at each. y-intercept: f(0) = -(1)(-2)(-4) = -8. Expanding, the leading term is -x³ (odd degree, negative coefficient), so the graph rises to the left and falls to the right: as x → -∞, f(x) → ∞, and as x → ∞, f(x) → -∞.
Question 18 of 20 · Short Answer
(+) Graph r(x) = (x² - 9)/(x² - x - 2). Factor first, then give the zeros, the vertical and horizontal asymptotes and the y-intercept.
r(x) = (x - 3)(x + 3)/((x - 2)(x + 1)). Zeros: 3 and -3. Vertical asymptotes: x = 2 and x = -1 (no factor cancels). Numerator and denominator both have degree 2 with leading coefficients 1, so the horizontal asymptote is y = 1. y-intercept: r(0) = -9/(-2) = 9/2. The graph crosses its horizontal asymptote once, where x² - 9 = x² - x - 2, at x = 7.
Question 19 of 20 · Short Answer
The polynomial f(x) = x³ - 3x + 1 has no rational zeros. Use technology to graph it, give its zeros to two decimal places, and state its end behavior. Why is technology the right tool here?
The graph crosses the x-axis three times, at about x ≈ -1.88, x ≈ 0.35 and x ≈ 1.53. It has a local maximum at (-1, 3) and a local minimum at (1, -1). The leading term x³ gives the end behavior: falls to the left and rises to the right. Testing ±1 (the only possible rational zeros) gives f(1) = -1 and f(-1) = 3, so no factorization over the integers is available and the zeros cannot be found by hand with the methods of this course.
Question 20 of 20 · Short Answer
Graph f(x) = 8(1/2)x. Give its intercepts and describe its end behavior on both sides.
y-intercept: f(0) = 8. Since 8(1/2)x > 0 for every x, there is no x-intercept. As x → ∞, f(x) → 0, so y = 0 is a horizontal asymptote on the right; as x → -∞, f(x) → ∞. Useful points: (1, 4), (2, 2), (3, 1), (-1, 16).
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSF.IF.C.7 mean?
HSF.IF.C.7 means students can graph a function from its equation and show its key features. The five parts list the families: linear and quadratic (a); square root, cube root, piecewise, step and absolute value (b); polynomial (c); rational, an advanced (+) part (d); and exponential, logarithmic and trigonometric (e). The main clause adds that simple cases are graphed by hand and complicated cases with technology.
Is HSF.IF.C.7 Algebra 1 or Algebra 2?
Both, and Precalculus as well. Parts a and b, and the exponential graphs in part e, are usually taught in Algebra I. Polynomial, logarithmic and trigonometric graphs are usually taught in Algebra II, and rational functions (part d, marked (+)) in Algebra II or Precalculus. The skill of reading features from an equation carries across all of them.
What counts as a key feature of a graph?
A key feature is a point or behavior that describes the graph: intercepts, maximum and minimum points, zeros, the starting point of a root graph, open and closed endpoints of a piecewise graph, asymptotes, end behavior, and the period, midline and amplitude of a trig graph. HSF.IF.B.4 asks students to interpret these features in context; HSF.IF.C.7 asks them to find and show them on a graph they draw.
When should students graph by hand and when with technology?
By hand when the equation reveals its features directly: a factored polynomial, a quadratic in vertex form, a shifted root or absolute value function. With technology when it does not, for example a cubic with no rational zeros such as x³ + x - 3. Even then, students should predict the end behavior and the number of possible turning points before they look at the screen, so they can spot a misleading window.
How do you find the end behavior of a polynomial?
Look only at the leading term. If its degree is even, both ends go the same way: up when the coefficient is positive, down when it is negative. If the degree is odd, the ends go opposite ways: a positive coefficient falls to the left and rises to the right, and a negative one does the reverse.
Why does a double zero make the graph touch the x-axis instead of crossing it?
A factor like (x + 3)² is never negative, so it does not change sign as x passes -3. The other factors keep their signs near x = -3, so the whole function keeps its sign on both sides: the graph touches the axis and turns back. A factor with an odd power, like (x + 3) or (x + 3)³, changes sign, so the graph crosses.
What are common mistakes when graphing piecewise and step functions?
Common ones: drawing each piece over the whole x-axis instead of only its interval, putting the open circle on the wrong piece, joining the steps of a step function with slanted or vertical lines, and confusing the ceiling function (rounds up, as in parking fees) with the floor function (rounds down). A quick check: every x-value should have exactly one filled point above or below it.
Do students need part d of HSF.IF.C.7 if they are not taking Precalculus?
No. Part d is marked (+), which Common Core uses for additional mathematics that students should learn to take advanced courses such as calculus. Many Algebra II courses include simple rational graphs anyway, because vertical and horizontal asymptotes come up again with exponential and logarithmic functions.
How do you find the period, midline and amplitude from an equation?
Write the function as y = A sin(Bx) + D or y = A cos(Bx) + D. The amplitude is |A|, the midline is y = D and the period is 2π/|B|. For y = 5 cos(πx/2) - 1, the amplitude is 5, the midline is y = -1 and the period is 2π/(π/2) = 4, so the graph runs between -6 and 4.
How does HSF.IF.C.7 connect to later courses?
Graphing from features is the starting point for transformations (HSF.BF.B.3), for modeling periodic data with trigonometric functions (HSF.TF.B.5), and for calculus, where students justify the same features (zeros, extrema, asymptotes, end behavior) with limits and derivatives.
07
Related Standards
6 standards
These standards connect to HSF.IF.C.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.F.A.3Prerequisite
Interpret y = mx + b as a linear function whose graph is a straight line
