Nonlinear function features and transformations
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Learning objectives
Connect an equation, a table, a graph, and an input-output statement without changing their meaning.
Function notation describes inputs and outputs
The statement f(a)
Evaluate a function by substituting the input everywhere the variable occurs, using parentheses around negative or compound inputs. Find an input for an output by setting the function equal to the given output and solving, with domain checks. One output may come from more than one input even though each allowed input has exactly one output.
Intercepts: solve f(x)
Domain and range: the domain is the allowed set of inputs; the range is the set of outputs actually attained. A formula may impose a restriction, and a context may impose additional ones. Do not confuse a finite graph window with the function’s full domain or range.
Recognize the basic function families
Parent rule | Domain; range | Recognizable feature |
|---|---|---|
x2 | R; [0, ∞) | Upward parabola, vertex (0, 0). |
x3 | R; R | Increasing curve through the origin. |
|x| | R; [0, ∞) | V-shape, vertex (0, 0). |
[0, ∞); [0, ∞) | Endpoint (0, 0), increasing to the right. | |
1/x | R ∖ {0}; R ∖ {0} | Two branches; axes are asymptotes. |
bx | R; (0, ∞) | Passes through (0, 1); b > 0, b ≠ 1. |
For an exponential, b > 1 gives increasing outputs and 0 < b < 1 gives decreasing outputs. Multipliers, translations, or domain restrictions change these basic features in systematic ways. The relevant question is not just “Which family?” but “Which transformed member of that family?”
Read a graph in a useful order
Inspect labeled points, intercepts, a vertex or endpoint, and any asymptotes. Check the scale of both axes. Then identify the simplest family and form consistent with those features. A curve’s apparent steepness on a stretched picture is less reliable than its exact labeled coordinates.
Transform points by matching the inside input
For g(x)
a point (u, v) on y
(h + , Av + k)
on y
New rule | What happens to the point (u, v) |
|---|---|
f(x − h) | (u + h, v): shift right by h when h > 0. |
f(x) + k | (u, v + k): shift up by k when k > 0. |
Af(x) | (u, Av): scale outputs; reflect across the x-axis if A < 0. |
f(Bx) | (u/B, v): scale horizontal distances by 1/|B|; reflect across the y-axis if B < 0. |
f(Bx + C) | ((u − C)/B, v): solve the inside equation directly. |
Inside operations act on inputs; outside operations act on outputs. The “opposite sign inside” shorthand works for a simple shift, but it is less safe than matching the old input. In f(2x−6)
Order matters. 2f(x) + 3 and 2(f(x) + 3) are different functions. The first doubles the old output and then adds 3; the second adds 3 first and doubles the result. Read parentheses as instructions.
A zero can move differently from a general point
An outside vertical shift can change which inputs are roots. Knowing f(r)
Unknown parameters. Use a vertex, endpoint, intercept, or known point to choose the form before solving for constants. Three expanded coefficients are often unnecessary when two graph features already determine the center and vertical offset.
Tables, repeated roots, and restricted ranges
For equally spaced inputs, a linear rule has constant first differences; a quadratic has constant second differences. For ax2 +bx+c sampled one unit apart, the second difference is 2a. For a pure exponential with positive values, equal input steps create constant ratios. These patterns are useful under a stated model assumption; a finite table by itself does not uniquely identify an arbitrary function.
Repeated roots leave a graphical clue. Near a polynomial root of even multiplicity, the graph touches the x-axis and turns back rather than changing sign. Near a root of odd multiplicity, it crosses. Use a known degree and another point to determine the scale; intercept locations alone may not uniquely determine a polynomial.
A restricted domain needs a fresh range analysis. For a quadratic on a closed interval, evaluate both endpoints and the vertex if its input is inside the interval. For a monotonic function, endpoint outputs determine the attained extremes; open endpoints may mean a bound is approached but not attained. A horizontal asymptote is not automatically an output the function can reach.
Use function structure before solving for everything
A nonlinear formula can produce a simpler relationship when two outputs are compared. In a difference f(x + h) − f(x) for a quadratic, the x2 terms cancel. In a ratio of outputs of a pure exponential, the initial coefficient cancels. In a rational transformation, rewriting a/(x − h) + k reveals the excluded input and offset immediately.
A compound input can also simplify a function. If its vertex is at x
Questions that connect representations
Can you turn f(a)
Calculator judgment. A table is useful for checking a proposed function at given inputs, and a graph is useful for comparing shape and intercepts. Neither a handful of matching values nor an attractive curve proves an identity. Use the algebra to establish exact equivalence, unknown-parameter conditions, and the number of possible inputs.
15 worked examples
6.01. Substitute a negative input with parentheses
If f(x) = x2 − 3x + 1, what is f(−2)?
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Recognize the structure. Replace every occurrence of x with −2, preserving the signs in the formula.
Work it through. f(−2) = (−2)2 − 3(−2) + 1 = 4 + 6 + 1 = 11.
Answer: 11.
Check. The corresponding graph point is (−2, 11): the negative number is the input, not necessarily the output.
Avoid the trap. (−2)2
6.02. Tell roots from a vertical intercept
For f(x) = (x − 4)(x + 1), what are the x-intercepts and the y-intercept of its graph?
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Recognize the structure. Zeros come from the factors; the vertical intercept comes from input 0.
Work it through. The equation f(x)
f(0) = (−4)(1) = −4.
Answer: x-intercepts (4, 0) and (−1, 0); y-intercept (0, −4).
Check. Each horizontal intercept has output zero, and the vertical intercept has input zero.
Avoid the trap. The constants inside the factors are not the two intercept coordinates. Solve each factor equal to zero, and do not reverse the order of an ordered pair.
6.03. Use symmetry in a table to build a quadratic
The function f is quadratic. Determine its rule from the table.
x
−1
0
1
2
f(x)
6
3
2
3
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Recognize the structure. Equal outputs at inputs 0 and 2 place the axis halfway between them.
Work it through. The axis is x
Answer: f(x)
Check. At x
Avoid the trap. The smallest displayed output is not automatically a global minimum for an arbitrary function. Here the stated quadratic model and symmetry justify the vertex conclusion.
6.04. Match a cubic graph to factors
A cubic polynomial has zeros at −2 and 1, touches the x-axis at 1, and passes through (0, −2), as shown. Write its rule.
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Recognize the structure. The touching root has even multiplicity; the degree is exactly three.
Work it through. The factors must be f(x)
Answer: f(x)
Avoid the trap. Using only one factor of x − 1 would create a crossing there and would not produce a cubic with the stated behavior.
6.05. Move a known point under two translations
The graph of f contains (−1, 4). If g(x) = f(x − 3) + 2, which corresponding point must lie on the graph of g?
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Recognize the structure. Make the new inside input equal to the old input −1.
Work it through. Solve x − 3
g(2) = f(−1) + 2 = 4 + 2 = 6.
Answer: (2, 6).
Check. The input shifted right by 3, from −1 to 2, and the output shifted up by 2, from 4 to 6.
Avoid the trap. Subtracting 3 from the old input would move the point left. Matching the inside input avoids the common reversed-shift error.
6.06. Apply outside operations in the stated order
If f(3) = −4 and h(x) = −2f(x) + 5, what is h(3)?
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Recognize the structure. The input remains 3; only the known output changes.
Work it through. h(3) = −2f(3) + 5 = −2(−4) + 5 = 13.
The factor −2 both reflects and scales the output before the vertical shift of 5.
Answer: 13.
Check. The point (3, −4) maps to (3, 13). Its horizontal coordinate does not change.
Avoid the trap. −2(f(3) + 5) would give a different result because it shifts before scaling. Follow the parentheses in the actual definition.
6.07. Handle an inside scale and shift together
Suppose f(4) = −2 and g(x) = f(2x − 6). Which corresponding point is guaranteed to be on the graph of g?
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Recognize the structure. The new inside expression must equal 4, the input for which the output is known.
Work it through. Solve 2x − 6
g(5) = f(4) = −2.
Answer: (5, −2).
Check. Writing 2x − 6
Avoid the trap. The inside constant −6 is not a simple right shift by 6 when the input also has coefficient 2. Solve the inside equation instead of guessing.
6.08. Reflect across both axes
The graph of f contains (−3, 5). If g(x) = −f(−x), which corresponding point is on the graph of g?
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Recognize the structure. The inside negative changes the input’s sign; the outside negative changes the output’s sign.
Work it through. To make −x
g(3) = −f(−3) = −5.
Answer: (3, −5).
Check. The original point first reflects across the y-axis to (3, 5) and then across the x-axis to (3, −5).
Avoid the trap. An inside negative alone does not change an output’s sign. The two minus signs act in different places and cannot simply be canceled.
6.09. Find a radical function’s domain
What is the greatest allowed real input of f(x) = ?
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Recognize the structure. The square-root radicand must be nonnegative.
Work it through.
7 − 2x ≥ 0 −2x ≥ −7 x ≤ .
The endpoint is included because a square root of zero is defined.
Answer: 7/2.
Check. f(7/2)
Avoid the trap. Dividing an inequality by −2 reverses its direction. Also, this is a greatest input; the outputs are not bounded above on the full domain.
6.10. Find a range on a restricted interval
Let f(x) = 4 − 2(x + 3)2 with domain −5 ≤ x ≤ 0. What is its range?
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Recognize the structure. Check the vertex and both endpoints of the allowed interval.
Work it through. The vertex is (−3, 4), inside the domain. Because the parabola opens downward, 4 is the maximum. At the endpoints,
f(−5) = 4 − 2(4) = −4, f(0) = 4 − 2(9) = −14.
The minimum on the interval is −14. The continuous quadratic attains every value between the minimum and maximum.
Answer: [−14, 4].
Check. Both bounds are attained: −14 at x
Avoid the trap. The unrestricted range would extend to negative infinity, but the stated domain cuts off those inputs. A range must reflect the actual domain.
6.11. Use a rational function’s structure to find a zero
The function f is defined by f(x) = + 4. What is the x-coordinate of its x-intercept?
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Recognize the structure. The intercept requires output zero, not input zero. The domain excludes x
Work it through. Set the function equal to zero:
= −4 3 = −4(x − 2).
Thus 3
Answer: 5/4.
Check. 5/4 − 2
Avoid the trap. The values 2 and 4 describe the vertical and horizontal asymptotes, not the intercepts. A visible parameter is meaningful, but not necessarily the quantity asked for.
6.12. Let a difference of outputs cancel the quadratic terms
Let f(x) = x2 − 4x + 7. If f(x + 2) − f(x) = 20, what is x?
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Recognize the structure. Substitute the compound input with parentheses; the leading squared terms will cancel.
Work it through. First, f(x + 2) = (x + 2)2 − 4(x + 2) + 7 = x2 + 3.
Therefore f(x + 2) − f(x) = x2 + 3 − (x2 − 4x + 7) = 4x − 4.
Set 4x − 4
Answer: 6.
Check. f(8)
Avoid the trap. f(x + 2) is not f(x) + 2. Also, the subtraction applies to all terms of f(x), not only its leading term.
6.13. Recover transformation parameters from graph features
Let f(x) = x2 and g(x) = af(x−h)+k, where a ≠ 0. The graph of g has vertex (4, −7) and passes through (6, 5). What is a + h + k?
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Recognize the structure. The vertex determines the horizontal and vertical shifts before the other point determines the scale.
Work it through. Since g(x)
5 = a(6 − 4)2 − 7 12 = 4a a = 3.
Then a + h + k = 3 + 4 − 7 = 0.
Answer: 0.
Check. g(6) = 3(2)2 − 7 = 5, and g(4)
Avoid the trap. The vertex input is h
6.14. Center a compound input around a quadratic’s vertex
Let f(x) = x2 − 4x + 1. What are all real values of a for which f(a + 2) = 10?
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Recognize the structure. The vertex form f(x)
Work it through. Substitute into the centered form:
f(a + 2) = ((a + 2) − 2)2 − 3 = a2 − 3.
Thus a2 − 3
Answer: a
Check. Both inputs 2 − and 2 + lie the same distance from the vertex input 2, and both give output 10.
Avoid the trap. The question asks for a, not the actual function input a + 2. Do not add 2 to the final answer unless that input is requested.
6.15. Use equal outputs and a minimum to avoid three equations
A quadratic f(x) = ax2 + bx + c, with a > 0, satisfies f(1) = f(7) = 20. Its minimum value is 2. What is c?
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Recognize the structure. Equal outputs at distinct inputs locate the axis of symmetry midway between them.
Work it through. The axis is x = (1 + 7)/2 = 4. The minimum gives vertex (4, 2), so
f(x)
Use f(1)
c = 2(0 − 4)2 + 2 = 34.
Answer: 34.
Check. The resulting rule 2(x − 4)2 + 2 gives 20 at both 1 and 7, and has minimum 2. Expanding gives 2x2 − 16x + 34.
Avoid the trap. The minimum 2 is the constant in vertex form, not the constant c in standard form. Equivalent forms display different features.
What this section should change in your approach
Translate notation into points, and transform a point by matching its old input. Use the whole domain when discussing a range or an extreme value. When parameters are unknown, graph features may identify the right form before any system of equations is necessary.
Section exit check
Attempt these five questions without the lesson or worked solutions. Give exact answers unless a decimal is requested. Open each model answer after attempting the question. These are learning checks, not an official score scale.
Mastery check
Explain Question 2 by solving for the inside input. For Question 4, list the vertex and endpoint values before giving the range. In Question 5, distinguish the scale parameter from the vertex coordinates.