High school Β· MathematicsAbout 120 minutes
MathematicsHigh schoolAbout 120 minutes

Graph polynomial, exponential, logarithmic, rational and trigonometric functions and indicate zeros, asymptotes, intercepts, end behavior, period, midline and amplitude

Several sessions: instruction, modeled examples, independent practice and a written explanation. Every worked example is meant to be re-checked by substitution before it is taught.

Ohio learning standards this draft targets

math-2017:F.IF.7math-2017:F.IF.7dmath-2017:F.IF.7fmath-2017:F.IF.7gmath-2017:F.IF.7h

Read the standard text

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.7d Graph polynomial functions, identifying zeros, when factoring is reasonable, and indicating end behavior. (A2, M3)

math-2017:F.IF.7f Graph exponential functions, indicating intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude. (A2, M3)

math-2017:F.IF.7g Graph rational functions, identifying zeros and asymptotes when factoring is reasonable, and indicating end behavior. (A2, M3)

math-2017:F.IF.7h Graph logarithmic functions, indicating intercepts and end behavior.

Provisional targets. Clause-by-clause educational review is still pending.

Section 1 Β· LearnLearn the idea
The cubic p(x) = xΒ³ βˆ’ 4x factors as x(x βˆ’ 2)(x + 2), so its zeros are βˆ’2, 0 and 2. It passes through (βˆ’1, 3) and (1, βˆ’3), is above the axis on βˆ’2 < x < 0 and below on 0 < x < 2, and because its degree is odd with a positive leading coefficient it falls without bound to the left and rises without bound to the right, reaching βˆ’5.625 at x = βˆ’2.5 and 5.625 at x = 2.5. Between the zeros it turns at a relative maximum of about 3.08 near x = βˆ’1.15 and a relative minimum of about βˆ’3.08 near x = 1.15 (exactly Β±16√3/9 at x = βˆ“2√3/3), not at the plotted points (βˆ’1, 3) and (1, βˆ’3).The cubic p(x) = xΒ³ βˆ’ 4x factors as x(x βˆ’ 2)(x + 2), so its zeros are βˆ’2, 0 and 2. It passes through (βˆ’1, 3) and (1, βˆ’3), is above the axis on βˆ’2 < x < 0 and below on 0 < x < 2, and because its degree is odd with a positive leading coefficient it falls without bound to the left and rises without bound to the right, reaching βˆ’5.625 at x = βˆ’2.5 and 5.625 at x = 2.5. Between the zeros it turns at a relative maximum of about 3.08 near x = βˆ’1.15 and a relative minimum of about βˆ’3.08 near x = 1.15 (exactly Β±16√3/9 at x = βˆ“2√3/3), not at the plotted points (βˆ’1, 3) and (1, βˆ’3).p(x) = xΒ³ βˆ’ 4x = x(x βˆ’ 2)(x + 2)-3-2-10123-6-4-20246(βˆ’2, 0)(0, 0)(2, 0)(βˆ’1, 3)(1, βˆ’3)xp(x)
The cubic p(x) = xΒ³ βˆ’ 4x factors as x(x βˆ’ 2)(x + 2), so its zeros are βˆ’2, 0 and 2. It passes through (βˆ’1, 3) and (1, βˆ’3), is above the axis on βˆ’2 < x < 0 and below on 0 < x < 2, and because its degree is odd with a positive leading coefficient it falls without bound to the left and rises without bound to the right, reaching βˆ’5.625 at x = βˆ’2.5 and 5.625 at x = 2.5. Between the zeros it turns at a relative maximum of about 3.08 near x = βˆ’1.15 and a relative minimum of about βˆ’3.08 near x = 1.15 (exactly Β±16√3/9 at x = βˆ“2√3/3), not at the plotted points (βˆ’1, 3) and (1, βˆ’3).

Learning goal: Graph polynomial functions, identifying zeros when factoring is reasonable and indicating end behavior; graph exponential and logarithmic functions, indicating intercepts and end behavior; graph rational functions, identifying zeros and asymptotes when factoring is reasonable and indicating end behavior; and graph trigonometric functions, showing period, midline and amplitude. Simple cases are graphed by hand and more complicated cases with technology, always checking the picture against the features the formula predicts.

Scope note. Parts g and h of F.IF.7 (graphing rational functions and graphing logarithmic functions) are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

Polynomial functions: zeros and end behavior. A polynomial function is a sum of terms with whole-number powers, and its graph is a smooth curve with no breaks, no asymptotes and no sharp corners. Two features give the main outline of its shape, and a few computed values fill in the rest. Its zeros are the inputs where the output is 0, the horizontal intercepts; they are found by factoring when factoring is reasonable. Its end behavior is decided entirely by the leading term, the term with the highest power, because for large |x| that term dwarfs the others. If the degree is odd and the leading coefficient positive, the graph falls without bound on the left and rises without bound on the right, like y = xΒ³; a negative leading coefficient reverses this. If the degree is even and the leading coefficient positive, the graph rises without bound on both sides, like y = xΒ²; a negative leading coefficient makes it fall on both sides.

A fully worked model: graph p(x) = xΒ³ βˆ’ 4x. Factor: x(xΒ² βˆ’ 4) = x(x βˆ’ 2)(x + 2), so the zeros are x = βˆ’2, 0 and 2, and the vertical intercept is p(0) = 0. The degree is 3, odd, with leading coefficient 1, so the graph falls to the left and rises to the right. Between consecutive zeros the sign is constant: p(βˆ’1) = βˆ’1 + 4 = 3 > 0, so the graph is above the axis on βˆ’2 < x < 0; p(1) = 1 βˆ’ 4 = βˆ’3 < 0, so it is below on 0 < x < 2. Values p(βˆ’1.5) = 2.625 and p(1.5) = βˆ’2.625 round out the bumps, and p(2.5) = 5.625 shows the rise. The graph has a relative maximum somewhere between βˆ’2 and 0 and a relative minimum between 0 and 2; technology locates them at about (βˆ’1.15, 3.08) and (1.15, βˆ’3.08). A factor that appears twice, as in (x βˆ’ 1)Β²(x + 2), gives a zero where the graph touches the axis and turns back without crossing, at x = 1, and crosses at x = βˆ’2. For q(x) = x⁴ βˆ’ 5xΒ² + 4 = (xΒ² βˆ’ 1)(xΒ² βˆ’ 4) = (x βˆ’ 1)(x + 1)(x βˆ’ 2)(x + 2), the four zeros are Β±1 and Β±2, the vertical intercept is 4, and the even degree with positive leading coefficient sends both ends upward.

Exponential and logarithmic functions. The exponential function y = 2Λ£ has vertical intercept (0, 1), no horizontal intercept, a horizontal asymptote y = 0 on the left, and unbounded growth on the right. The logarithmic function y = logβ‚‚x is its inverse: it undoes 2Λ£, so its graph is the reflection of y = 2Λ£ across the line y = x. Every feature reflects. Its domain is x > 0, its horizontal intercept is (1, 0) because logβ‚‚1 = 0, and it has no vertical intercept; the vertical line x = 0 is a vertical asymptote, and as x decreases toward 0 the outputs fall without bound: logβ‚‚(1/2) = βˆ’1, logβ‚‚(1/4) = βˆ’2. As x increases the outputs rise without bound but very slowly: logβ‚‚2 = 1, logβ‚‚4 = 2, logβ‚‚8 = 3. The same shape holds for log₁₀x and for the natural logarithm ln x, with intercept (1, 0) in every base. A general exponential function y = a Β· bΛ£ has intercept (0, a); when a > 0, it increases for b > 1 and decreases for 0 < b < 1, with the asymptote on the right in the second case. A negative a reflects the graph across the horizontal axis and reverses the direction: βˆ’2(1/2)Λ£ increases.

Rational functions. A rational function is a quotient of polynomials, such as r(x) = (x βˆ’ 1)/(x βˆ’ 3). Its zeros are the zeros of the numerator that are not also zeros of the denominator: here x = 1. Its vertical asymptotes are at zeros of the denominator that are not canceled by a common factor with the numerator (cancel every common factor first: a zero of the original denominator that no longer appears in the reduced denominator leaves a hole, while one that still appears, as for (x βˆ’ 1)/(x βˆ’ 1)Β², is still a vertical asymptote), here x = 3, where the function is undefined and the outputs grow without bound in size as x approaches 3: r(2) = βˆ’1, r(2.9) = βˆ’19, r(3.1) = 21, r(4) = 3. Its end behavior comes from comparing degrees: the numerator and denominator both have degree 1 with leading coefficients 1 and 1, so as x grows in either direction r(x) approaches 1/1 = 1, and y = 1 is a horizontal asymptote. Check: r(5) = 2, r(7) = 1.5, r(0) = 1/3. The graph has two separate branches, one on each side of x = 3. Factoring is reasonable when the zeros of numerator and denominator are integers; otherwise technology finds them.

Trigonometric functions. The sine and cosine functions are periodic: their outputs repeat every 2Ο€ radians. The basic graph y = sin x oscillates between βˆ’1 and 1, crossing the horizontal axis at 0, Ο€, 2Ο€ and so on. Three numbers describe any graph of the form y = A sin(Bx) + D or y = A cos(Bx) + D. The midline is the horizontal line y = D halfway between the highest and lowest outputs. The amplitude is |A|, the distance from the midline to the maximum or minimum. The period is 2Ο€/|B| (for B β‰  0), the length of input over which the graph completes one cycle. For y = 3 sin x + 1 the midline is y = 1, the amplitude is 3, so the graph runs from 1 βˆ’ 3 = βˆ’2 up to 1 + 3 = 4, and the period is 2Ο€. For y = 2 cos(Ο€x/4) the midline is y = 0, the amplitude is 2 and the period is 2Ο€/(Ο€/4) = 8; the graph starts at its maximum (0, 2), falls to its minimum (4, βˆ’2) and returns to (8, 2). To graph, draw the midline, mark the maximum and minimum heights, divide one period into four equal parts and plot the five key points, then repeat the cycle.

Applications: letting the situation choose the family. The families differ in features that situations also have, so the features justify the choice of model. The following situations are hypothetical. An open-top box is folded from a 20 cm square sheet by cutting squares of side x from the corners; its volume V(x) = x(20 βˆ’ 2x)Β² is a polynomial because a volume is a product of three lengths, and its zeros x = 0 and x = 10 mark where the box collapses, so they bound the domain 0 < x < 10. Technology locates the relative maximum at x = 10/3 β‰ˆ 3.33 cm, where V = 16,000/27 β‰ˆ 592.6 cmΒ³. A hypothetical print shop pays a 500-dollar setup fee plus 2 dollars per poster, so the average cost per poster is A(n) = (500 + 2n)/n, a rational function: its vertical asymptote at n = 0 lies outside the domain n β‰₯ 1 but explains why a handful of posters is very expensive each, and its horizontal asymptote A = 2 is the per-poster cost that the average approaches, A(100) = 7 and A(1,000) = 2.5 dollars. A quantity that halves every fixed time, such as a medicine level, is exponential because its percent change is constant, and a quantity that repeats every day, such as the height of the tide, is trigonometric because it is periodic. In each case a feature of the graph, a bounded domain, a leveling-off value, a constant ratio or a repeating cycle, matches a characteristic of the situation, and that match is the reason for the model.

Reading a technology graph. Rational and trigonometric graphs are the ones most often drawn with technology, and they are the ones most easily misread. A calculator may draw a nearly vertical line across a vertical asymptote that is not part of the graph, and a sine curve sampled at too few points looks jagged or even appears to have the wrong period. The defense is the same as always: compute the features from the formula first, then confirm that the picture shows them.

Section 2 Β· ModelWork through the examples

Worked polynomial graph: Graph q(x) = x⁴ βˆ’ 5xΒ² + 4. Factor as (xΒ² βˆ’ 1)(xΒ² βˆ’ 4) = (x βˆ’ 1)(x + 1)(x βˆ’ 2)(x + 2), so the zeros are βˆ’2, βˆ’1, 1 and 2. Vertical intercept: q(0) = 4. End behavior: degree 4, leading coefficient positive, so both ends rise. Signs between zeros: q(1.5) = 5.0625 βˆ’ 11.25 + 4 = βˆ’2.1875 < 0, so the graph dips below the axis between 1 and 2, and by symmetry between βˆ’2 and βˆ’1; q(0) = 4 > 0, so it is above the axis between βˆ’1 and 1. The curve therefore has a W shape: down through βˆ’2, up through βˆ’1 to a relative maximum of 4 at x = 0, down through 1, up through 2.

Worked rational graph: Graph r(x) = (x βˆ’ 1)/(x βˆ’ 3). Zero: numerator zero at x = 1, so (1, 0). Vertical intercept: r(0) = (βˆ’1)/(βˆ’3) = 1/3. Vertical asymptote: x = 3, from the denominator. Horizontal asymptote: the degrees are equal, so y = 1/1 = 1. Points: r(2) = βˆ’1, r(2.5) = βˆ’3, r(3.5) = 5, r(4) = 3, r(5) = 2, r(7) = 1.5. The left branch passes through (0, 1/3), (1, 0), (2, βˆ’1) and plunges downward as x approaches 3 from the left; the right branch comes down from very large values just right of 3, through (4, 3) and (5, 2), and flattens toward y = 1.

Worked trigonometric graph: Graph y = 3 sin x + 1 over one period. Midline: y = 1. Amplitude: 3, so the maximum is 4 and the minimum is βˆ’2. Period: 2Ο€. Key points, dividing the period into quarters: (0, 1) on the midline, (Ο€/2, 4) the maximum, (Ο€, 1) back on the midline, (3Ο€/2, βˆ’2) the minimum, and (2Ο€, 1). Draw the smooth wave through these five points and repeat it in both directions. The graph never has an end behavior of growing without bound; it stays between βˆ’2 and 4 forever.

Section 3 Β· DoPractice, demonstrate, and explain

Representation task: Graph p(x) = xΒ³ βˆ’ 4x by hand on axes for βˆ’3 ≀ x ≀ 3 and βˆ’6 ≀ y ≀ 6: factor to find and label the zeros, mark p(βˆ’1) = 3 and p(1) = βˆ’3, and write the end behavior at each end of the curve. Then use technology to graph the same function, locate its relative maximum and minimum to two decimal places, and record whether the picture agrees with every feature you predicted.

Independent performance: Produce a portfolio of five graphs, each with its features listed and verified: a factorable cubic or quartic polynomial with its zeros and end behavior; an exponential function with its intercept and asymptote; a logarithmic function with its intercept, asymptote and domain; a rational function whose numerator and denominator factor, with its zeros, its vertical asymptote and its horizontal asymptote (choose a numerator whose degree is no greater than the denominator's, so that a horizontal asymptote exists, or state that there is none); and a sine or cosine function of the form A sin(Bx) + D with its period, midline and amplitude. Graph the polynomial and the trigonometric function by hand and the others with technology, and for each technology graph state one feature you checked against the formula. For at least two of the five, describe a situation the function models and explain which characteristic of the situation (a bounded domain, a leveling-off value, a constant percent change or a repeating cycle) makes that family the appropriate choice. Success criteria: every zero and asymptote is found algebraically; end behavior is justified from the leading term or from the degrees; period, midline and amplitude are computed from A, B and D and match the drawing; and the domain of the logarithmic function is found from its formula (x > 0 for logβ‚‚x, but x > 2 for logβ‚‚(x βˆ’ 2)).

Section 4 Β· Check8-question lesson practice test

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.

Problem 1

Find the zeros of p(x) = xΒ³ βˆ’ 4x by factoring and state its end behavior with a reason.

Show worked solution

xΒ³ βˆ’ 4x = x(xΒ² βˆ’ 4) = x(x βˆ’ 2)(x + 2), so the zeros are x = 0, 2 and βˆ’2. The leading term xΒ³ has odd degree and positive coefficient, so as x increases without bound p(x) rises without bound, and as x decreases without bound p(x) falls without bound; the smaller term βˆ’4x cannot change this for large |x|.

Problem 2

Describe the graph of y = logβ‚‚x: its domain, intercepts, asymptote and end behavior, and explain how it is related to y = 2Λ£.

Show worked solution

The domain is x > 0. The horizontal intercept is (1, 0) because logβ‚‚1 = 0; there is no vertical intercept. The line x = 0 is a vertical asymptote: as x decreases toward 0 the outputs fall without bound, for example logβ‚‚(1/4) = βˆ’2. As x increases the outputs rise without bound but slowly, logβ‚‚8 = 3. The graph is the reflection of y = 2Λ£ across the line y = x because the two functions are inverses.

Problem 3

For r(x) = (x βˆ’ 1)/(x βˆ’ 3), identify the zero, the vertical asymptote and the horizontal asymptote, and give r(4).

Show worked solution

The numerator is zero at x = 1, so the zero is (1, 0). The denominator is zero at x = 3, so x = 3 is a vertical asymptote where the function is undefined. Numerator and denominator have equal degree with leading coefficients 1 and 1, so y = 1 is the horizontal asymptote. r(4) = 3/1 = 3.

Problem 4

State the period, midline and amplitude of y = 2 cos(Ο€x/4), and give its maximum and minimum values.

Show worked solution

Amplitude is |A| = 2, midline is y = 0 since D = 0, and period is 2Ο€/(Ο€/4) = 8. The maximum is 0 + 2 = 2, first at x = 0, and the minimum is 0 βˆ’ 2 = βˆ’2, at x = 4; the graph returns to its maximum at x = 8, completing one cycle.

Problem 5

The polynomial (x βˆ’ 1)Β²(x + 2) has zeros at 1 and βˆ’2. Explain how the graph behaves differently at the two zeros.

Show worked solution

At x = βˆ’2 the factor appears once, so the sign of the output changes and the graph crosses the axis. At x = 1 the factor (x βˆ’ 1)Β² is never negative, so the output does not change sign there; the graph touches the axis at (1, 0) and turns back without crossing. Testing x = 0 gives 1 Γ— 2 = 2 > 0 and x = 2 gives 1 Γ— 4 = 4 > 0, confirming the same sign on both sides of 1.

Problem 6

Graph y = 3 sin x + 1 over one period by listing its five key points, and explain why the function has no unbounded end behavior.

Show worked solution

Midline y = 1, amplitude 3, period 2Ο€. Key points: (0, 1), (Ο€/2, 4), (Ο€, 1), (3Ο€/2, βˆ’2), (2Ο€, 1). Because sin x always lies between βˆ’1 and 1, the output always lies between 1 βˆ’ 3 = βˆ’2 and 1 + 3 = 4; the graph repeats the same wave forever and never grows without bound in either direction.

Problem 7

Graphing technology draws r(x) = (x βˆ’ 1)/(x βˆ’ 3) with a steep line joining the two branches near x = 3. Is that line part of the graph? How do you know?

Show worked solution

No. The function is undefined at x = 3, so no point of the graph has x-coordinate 3, and the outputs near 3 go to very large negative values on the left and very large positive values on the right. The calculator merely connected consecutive sampled points across the asymptote. Knowing from the denominator that x = 3 is a vertical asymptote tells you to ignore the drawn line.

Problem 8

For q(x) = x⁴ βˆ’ 5xΒ² + 4, find all zeros and the vertical intercept, and state the end behavior.

Show worked solution

q(x) = (xΒ² βˆ’ 1)(xΒ² βˆ’ 4) = (x βˆ’ 1)(x + 1)(x βˆ’ 2)(x + 2), so the zeros are 1, βˆ’1, 2 and βˆ’2. The vertical intercept is q(0) = 4. The degree 4 is even and the leading coefficient positive, so the graph rises without bound at both ends, giving a W shape with the relative maximum 4 at x = 0.