math-2017:F.IF.7math-2017:F.IF.7c
math-2017:F.IF.7 Graph functions expressed symbolically and indicate key features of the graph, by hand in simple cases and using technology for more complicated cases. Include applications and how key features relate to characteristics of a situation, making selection of a particular type of function model appropriate.
math-2017:F.IF.7c Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. (A2, M3)
Learning goal: Graph square root functions, cube root functions and piecewise-defined functions, including step functions and absolute value functions, from their formulas, indicating the domain, intercepts, endpoints, turning points and end behavior, and connect the shape of each to the kind of situation it models.
Square root functions. The function f(x) = √x is defined only for x ≥ 0, because the square root of a negative number is not a real number. Its graph starts at the origin, the point (0, 0), which is both intercepts, and rises to the right forever, more and more slowly: √1 = 1, √4 = 2, √9 = 3, √16 = 4. The end behavior is that f(x) increases without bound as x increases, even though it slows; there is no left end, because the domain stops at 0. For a shifted version such as g(x) = √(x − 1) + 2, first find the domain from x − 1 ≥ 0, so x ≥ 1, and then the endpoint: g(1) = 2, so the graph starts at (1, 2). Points g(2) = 3, g(5) = 4 and g(10) = 5 fix the shape. The vertical intercept would need x = 0, which is outside the domain, so there is none; the horizontal intercept would need √(x − 1) = −2, impossible, so there is none either. A square root graph is half of a sideways parabola, and the endpoint is its most important feature.
Cube root functions. The function c(x) = ∛x is defined for every real number, because every real number, negative ones included, has exactly one real cube root: ∛(−8) = −2 since (−2)³ = −8, ∛(−1) = −1, ∛0 = 0, ∛1 = 1, ∛8 = 2. The graph passes through the origin, rises through the whole plane from lower left to upper right, is steepest at the origin and flattens out in both directions. Its end behavior is upward without bound to the right and downward without bound to the left. The contrast with the square root is the domain: a square root function has an endpoint where its domain begins; a cube root function has no endpoint at all. The graph of ∛x also has a symmetry, ∛(−x) = −∛x, which is a point symmetry through the origin, taken up again in the lesson on even and odd functions.
Piecewise-defined functions. A piecewise-defined function uses different formulas on different parts of its domain. A fully worked hypothetical model: a parking garage charges 3 dollars for the first hour or any part of it and then 2 dollars for each additional hour or part, up to a daily maximum of 15 dollars. Writing t for hours parked, the cost is C(t) = 3 for 0 < t ≤ 1, C(t) = 5 for 1 < t ≤ 2, C(t) = 7 for 2 < t ≤ 3, and so on, until the cost reaches 15 once t passes 6 hours and stays 15 for the rest of the day; the model covers one day of parking, 0 < t ≤ 24. To graph a piecewise function, graph each piece over its own interval only, marking each endpoint with a filled dot if the endpoint belongs to the piece and an open circle if it does not. Here the horizontal segment at height 5 runs from an open circle at t = 1 to a filled dot at t = 2, because parking exactly one hour costs 3, not 5. The function jumps at t = 1, 2, 3, 4, 5 and 6; after the cap is reached there are no more jumps. These jumps are the key feature of a step function: a function whose graph is a series of horizontal segments. Step functions model prices that change at thresholds, postage by weight bands, and any quantity that is rounded. The greatest integer function ⌊x⌋, the largest integer less than or equal to x, is the basic example: ⌊2.7⌋ = 2, ⌊3⌋ = 3 and ⌊−1.2⌋ = −2, with filled dots at each integer on the left of each step and open circles on the right.
By hand or with technology. The functions in this lesson are simple enough to graph by hand; more complicated ones are graphed with technology and then checked. Take f(x) = x² − 2 for x < 1, f(x) = 3 − 0.5x for 1 ≤ x ≤ 4, and f(x) = √(x − 4) + 1 for x > 4. A graphing tool draws the three pieces quickly, but it may join the pieces with a vertical segment and it does not show which endpoint is filled. So check each boundary by computing: at x = 1 the left piece approaches 1 − 2 = −1 (open circle at (1, −1)) while f(1) = 3 − 0.5 = 2.5 (filled dot), a jump of 3.5; at x = 4 the middle piece gives f(4) = 1 and the right piece starts at √0 + 1 = 1, so the pieces meet with no jump. The domain is all real numbers, the vertical intercept is f(0) = −2, and the only zero is x = −√2 ≈ −1.41, since x² − 2 = 0 at x = √2 lies outside x < 1, 3 − 0.5x = 0 at x = 6 lies outside 1 ≤ x ≤ 4, and the last piece is always at least 1.
A piecewise function need not be a step function. The function p(x) = x² for x < 1 and p(x) = 3 − x for x ≥ 1 has a parabola piece on the left and a line piece on the right. At x = 1 the parabola piece would give 1 and the line piece gives 2; only the line piece includes x = 1, so the graph has an open circle at (1, 1) and a filled dot at (1, 2), and the function jumps there. Had the second formula been 2 − x, both pieces would meet at (1, 1) and the graph would be connected. Checking whether the pieces meet at each boundary is the step students skip.
Absolute value functions. The absolute value function a(x) = |x| equals x when x ≥ 0 and −x when x < 0, so it is a piecewise function with two linear pieces meeting at the origin in a V shape. It is symmetric about the vertical axis and its minimum is 0 at the vertex (0, 0). A transformed version such as v(x) = |x − 3| − 2 has vertex at x = 3, since the inside is zero there, with v(3) = −2. The horizontal intercepts solve |x − 3| = 2, so x − 3 = 2 or x − 3 = −2, giving x = 5 and x = 1; the vertical intercept is v(0) = 3 − 2 = 1. Points v(−1) = 2 and v(7) = 2 show the two arms, each with slope ±1. The end behavior is upward without bound in both directions, and the function is decreasing for x < 3 and increasing for x > 3. Absolute value models distance from a target: |x − 3| is the distance of x from 3, so v measures how far a reading is from a target of 3 minus a tolerance of 2, and the intercepts 1 and 5 are the edges of the acceptable band.
Which feature belongs to which situation. A square root graph, starting at a point and rising with slowing growth, models quantities such as the time for a pendulum swing against its length, where a starting value exists and growth slows. A step function models thresholds and rounding. An absolute value function models distance from a target. A cube root graph has no endpoint, accepts negative as well as positive inputs, and grows in both directions ever more slowly; on its nonnegative part it models, for example, the edge length of a cube as a function of its volume. Selecting the model means matching these characteristics, and graphing the model means drawing its endpoint, its jumps or its vertex correctly and stating its domain.
Worked square root graph: Graph g(x) = √(x − 1) + 2. Domain: x ≥ 1. Endpoint: (1, 2), drawn as a filled dot. Points: (2, 3), (5, 4), (10, 5). No vertical intercept, because 0 is not in the domain; no horizontal intercept, because the smallest output is 2. The graph rises to the right without bound but ever more slowly: from x = 1 to x = 2 it rises by 1, from x = 2 to x = 5 by another 1, and from x = 5 to x = 10 by another 1. The range is y ≥ 2.
Worked piecewise graph with a jump check: Graph p(x) = x² for x < 1 and p(x) = 3 − x for x ≥ 1. Left piece: the parabola through (−2, 4), (−1, 1), (0, 0), ending at an open circle at (1, 1), because x = 1 is not in this piece. Right piece: the line through a filled dot at (1, 2), then (2, 1), (3, 0) and (5, −2). The two pieces do not meet, so the graph jumps from 1 up to 2 at x = 1. The horizontal intercepts are (0, 0), from the left piece, and (3, 0), from the right piece; the vertical intercept is (0, 0). End behavior: upward without bound on the left, from the parabola, and downward without bound on the right, from the line with slope −1.
Worked absolute value graph: Graph v(x) = |x − 3| − 2. The vertex is where the inside is zero, at x = 3, with v(3) = −2. Horizontal intercepts: |x − 3| = 2 gives x = 1 and x = 5. Vertical intercept: v(0) = 3 − 2 = 1. Arms: v(−1) = 2 and v(7) = 2. Draw the V with vertex (3, −2), through (1, 0) and (5, 0), rising with slope −1 to the left and +1 to the right. The minimum is −2, the function is decreasing for x < 3 and increasing for x > 3, and the graph is symmetric about the line x = 3.
Representation task: On one set of axes for −2 ≤ x ≤ 10, graph v(x) = |x − 3| − 2 and g(x) = √(x − 1) + 2. Label the vertex of v, both of its horizontal intercepts, its vertical intercept, and the endpoint of g. Then make a table of the parking cost C(t) for t = 0.5, 1, 1.5, 2, 2.5, 3, 7, 8 and 10 hours, graph C as a step function with correct filled and open endpoints, and write a sentence explaining why the dot at t = 1 is at height 3 and not 5.
Independent performance: Write and graph four functions: a square root function with a shifted endpoint, a cube root function, a piecewise-defined function with at least two non-constant pieces, and an absolute value function with a vertex away from the origin. For each, state the domain, every intercept, any endpoint, vertex or jump, and the end behavior. For the piecewise function, check at each boundary whether the pieces meet and mark the endpoints correctly. Then describe one real situation each function could model and say which feature of the graph corresponds to which feature of the situation. Success criteria: domains are correct, in particular the square root domain comes from the radicand and the cube root domain is all real numbers; open and filled endpoints are used correctly; every intercept is verified algebraically; and the situation matches the shape, for example thresholds for a step function or distance from a target for an absolute value function.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Find the domain of g(x) = √(x − 1) + 2, its endpoint, and explain why it has neither a vertical nor a horizontal intercept.
The radicand must satisfy x − 1 ≥ 0, so the domain is x ≥ 1 and the graph starts at the endpoint (1, 2). A vertical intercept would require x = 0, which is outside the domain. A horizontal intercept would require √(x − 1) = −2, but a square root is never negative, so the output is always at least 2.
Explain why ∛x is defined for negative inputs but √x is not, and give ∛(−8).
Every real number has exactly one real cube root because a negative number cubed is negative: (−2)³ = −8, so ∛(−8) = −2. No real number squared is negative, so √(−8) has no real value. That is why the cube root graph is defined for every x, extending without bound to the left and right, while the square root graph has an endpoint at x = 0.
A garage charges 3 dollars for the first hour or part and 2 dollars for each additional hour or part, to a maximum of 15 dollars. Give C(1), C(1.5) and C(9), and say how the graph shows the jump at t = 1.
C(1) = 3, because exactly one hour is still the first hour. C(1.5) = 5, because any part of a second hour adds 2 dollars. C(9) = 15, the daily maximum: the charge first reaches 3 + 2 × 6 = 15 as soon as the stay passes 6 hours (6 < t ≤ 7) and is held there afterward. The graph shows the jump with a filled dot at (1, 3) ending the first step and an open circle at (1, 5) beginning the second, so t = 1 has the single output 3.
The function p(x) = x² for x < 1 and p(x) = 3 − x for x ≥ 1 is graphed. Describe what happens at x = 1 and explain how the endpoints are marked.
The left piece approaches the value 1² = 1 as x nears 1 but does not include x = 1, so it ends in an open circle at (1, 1). The right piece includes x = 1 and gives 3 − 1 = 2, so it starts with a filled dot at (1, 2). The two values differ, so the graph jumps from 1 to 2 at x = 1; the single output at x = 1 is 2.
For v(x) = |x − 3| − 2, find the vertex, the intercepts and the intervals on which the function is increasing and decreasing.
The inside x − 3 is zero at x = 3, and v(3) = −2, so the vertex is (3, −2), a minimum. Horizontal intercepts: |x − 3| = 2 gives x = 1 or x = 5. Vertical intercept: v(0) = |−3| − 2 = 1. The left arm has slope −1, so v is decreasing for x < 3; the right arm has slope +1, so v is increasing for x > 3.
Evaluate the greatest integer function at 2.7, 3 and −1.2, and explain why its graph is a step function.
⌊2.7⌋ = 2, ⌊3⌋ = 3 and ⌊−1.2⌋ = −2, the largest integer not exceeding each input; note that −2 is below −1.2, not −1. Between consecutive integers the output is constant, so the graph is a horizontal segment on each unit interval, filled at the left integer and open at the right, and the graph jumps up by 1 at every integer, which is the defining shape of a step function.
A machine should produce parts 3 cm long, and a part is rejected if its length differs from 3 cm by more than 2 cm. Explain how v(x) = |x − 3| − 2 models this and what the intercepts x = 1 and x = 5 mean.
|x − 3| is the distance of a measured length x from the target 3, and subtracting 2 compares that distance with the tolerance. The function is negative or zero when the part is within 2 cm of the target and positive when it is rejected. The intercepts x = 1 and x = 5 are the shortest and longest acceptable lengths, the edges of the acceptable band.
Graph f(x) = x² − 2 for x < 1, 3 − 0.5x for 1 ≤ x ≤ 4, and √(x − 4) + 1 for x > 4 with technology. What must you check by computation that the technology graph may not show, and what do you find at x = 1 and x = 4?
The technology graph may join pieces with a vertical segment and does not mark open or filled endpoints, so each boundary must be computed. At x = 1 the left piece approaches (1, −1), an open circle, while f(1) = 2.5 is a filled dot: a jump of 3.5. At x = 4 the middle piece gives f(4) = 1 and the right piece starts at 1, so the graph is unbroken there. The vertical intercept is f(0) = −2 and the only zero is x = −√2 ≈ −1.41.