Functions and Their Graphs
A complete learning module with worked examples, interactive graphs, domain practice, collapsible solutions, and selected review problems.
What is a function?
A function describes a dependable connection between an input and an output. Different inputs may lead to the same output; a single input cannot lead to two different outputs.
Definition 1 · Function, domain, and range
A function f: X → Y assigns exactly one element of Y to every element of X.
- Domain: all admissible input values.
- Range: all output values actually produced.
- Codomain: the target set Y; it may contain values that never occur as outputs.
We write y = f(x), read “y equals f of x.”
Example 1 · One rule, five representations
Consider f(x) = x². Here are five ways to describe it.
Mapping
−1 → 1
0 → 0
1 → 1
2 → 4
Table of values
| x | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| f(x) | 4 | 1 | 0 | 1 | 4 |
Ordered pairs
{(−2,4), (−1,1), (0,0), (1,1), (2,4)}
Equation
y = x²
The displayed sample has domain {−2,−1,0,1,2} and range {0,1,4}. With the unrestricted real rule f(x)=x², the domain is ℝ and range is [0,∞).
Common mistake · Repeated output ≠ repeated input
The pairs (−1,1) and (1,1) are allowed. The pairs (1,−1) and (1,1) together fail: the same input 1 has two outputs.
Try it 1 · Function or not?
A table includes (0,0), (1,−1), (1,1), (4,2), (4,−2). Does this relation define a function? Then decide for {(4,6),(6,4),(−1,1),(1,−1),(0,2),(2,4)}.
Show solution
The first relation is not a function: input 1 has outputs −1 and 1 (and input 4 also has two outputs). The second is a function: each input occurs only once. Its domain is {−1,0,1,2,4,6}; its range is {−1,1,2,4,6}.
Graph criterion · The vertical line test
A graph represents y as a function of x exactly when every vertical line meets the graph at most once. Zero intersections mean that input is outside the domain.
A single failing line disproves “function.” One passing line alone does not prove it.
Imagine giving a rule one input and asking it to return an output. The key question is whether that input determines just one result. Different inputs may share an output; the restriction works in the other direction. As we move between tables, diagrams, and graphs, look for that same idea in each representation.
Notation & evaluation
Definition 2 · Evaluating a function
To evaluate f at an input, replace every occurrence of the independent variable with that input.
Use parentheses when substituting a negative number or an expression, so the original operations apply to the entire input.
Example 2 · The function machine
f(x) = 2x² − 5
Input 2 → square → multiply by 2 → subtract 5 → output 3
f(2) = 2(2)² − 5 = 8 − 5 = 3
For an algebraic input:
f(a+1) = 2(a+1)² − 5 = 2(a²+2a+1) − 5 = 2a²+4a−3
Common mistake · f(x) is not f times x
The notation names an output. Also, (a+1)² = a²+2a+1, not a²+1. Keep the middle term when expanding a square.
Try it 2 · Evaluate and simplify
Given f(x)=3−5x and g(x)=x²+3x+1, find:
- f(3)
- f(x+7)
- g(−5)
- g(2m+1)
Show solution
- 3−5(3)=−12
- 3−5(x+7)=−5x−32
- (−5)²+3(−5)+1=25−15+1=11
- (2m+1)²+3(2m+1)+1=4m²+10m+5
Having decided that a rule is a function, we can ask for its output at a particular input. Function notation tells us which input to substitute; it does not tell us to multiply by the function name. Replace every occurrence of the input variable consistently, and preserve the grouping when the substituted value is negative.
Finding the domain
Unless a context or a stated set restricts the inputs, find the largest domain for which the expression is a real number.
Domain rules
| Expression | Restriction | Example |
|---|---|---|
| Polynomial | Every real input is allowed. | 5x+4: D=ℝ |
| Rational expression | Denominator ≠ 0. | 5x+44−3x: x ≠ 4/3 |
| Even root | Radicand ≥ 0. | √(4−3x): x ≤ 4/3 |
| Odd root | Every real radicand is allowed. | ∛(4−3x): D=ℝ |
| Even root in a denominator | Radicand > 0. | 1√(x−2): x > 2 |
Example 3 · Combine every restriction
Find the domain of h(x)=x+1√(x²+3x−40).
The square root is in the denominator, so x²+3x−40 > 0, not merely ≥ 0. Factor:
(x+8)(x−5) > 0
The product is positive outside its roots. Therefore D=(−∞,−8) ∪ (5,∞).
Reading intervals
A square bracket includes an endpoint; a parenthesis excludes it. Infinity always takes a parenthesis. The symbol ∪ joins allowed intervals; ℝ means all real numbers.
Try it 3 · Domain practice
- f(x)=x²+2x−8x²−9
- g(x)=3+√(5x+14)
- m(x)=6x+5⁵√(x+6)+4
- n(x)=3⁴√(x+6)+√(7−x)
- h(x)=4x²+√(x²−25x²−x−2)
Show solution
- x²−9=(x−3)(x+3) ≠ 0, so D=ℝ ∖ {−3,3}.
- 5x+14 ≥ 0, so D=[−14/5,∞).
- An odd root allows negative radicands, but this denominator cannot be zero. Thus D=ℝ ∖ {−6}.
- Both x+6 ≥ 0 and 7−x ≥ 0 must hold. Thus D=[−6,7]. The coefficient 3 does not change the restrictions.
- Require (x−5)(x+5)/[(x−2)(x+1)] ≥ 0, excluding −1 and 2. Testing the five intervals gives D=(−∞,−5] ∪ (−1,2) ∪ [5,∞).
We have been putting inputs into a rule and calculating outputs. Now reverse the question: which inputs are we allowed to use at all? A formula can look perfectly ordinary and still contain a division or a square root that restricts it. Checking this first prevents us from drawing points that do not belong to the function.
Basic function families
Recognizing the parent graph gives you a starting point. Transformations will move or reflect it.
Linear
f(x)=x
D=ℝ; R=ℝ. A line with slope 1.
Constant
f(x)=2
D=ℝ; R={2}. A horizontal line.
Quadratic
f(x)=x²
D=ℝ; R=[0,∞). Vertex (0,0).
Cubic polynomial
f(x)=x³
D=ℝ; R=ℝ. An odd-degree example.
Square root
f(x)=√x
D=[0,∞); R=[0,∞). Starts at (0,0).
Absolute value
f(x)=|x|
D=ℝ; R=[0,∞). A V-shaped graph.
Rational
f(x)=1/x
D=ℝ ∖ {0}; R=ℝ ∖ {0}. Two separate branches.
Definition 3 · Polynomial
A polynomial has the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ⋯ + a₁x + a₀, where the exponents are nonnegative integers and the coefficients are real. For a nonzero polynomial of degree n, aₙ ≠ 0.
Linear and quadratic functions are examples. Over the real numbers, polynomial functions have domain ℝ, but their ranges need not be ℝ: a square cannot produce a negative output.
Quadratic vertex and range
For f(x)=ax²+bx+c, a ≠ 0, the vertex has
h=−b2a, k=f(h)=4ac−b²4a.
If a>0, the parabola opens upward and R=[k,∞). If a<0, it opens downward and R=(−∞,k].
Try it 4 · Read the family
For q(x)=x²−3x+5, find the vertex, domain, and range. Is the vertical line x=2 a function of x?
Show solution
h=3/2 and q(3/2)=11/4. Vertex: (3/2,11/4); domain: ℝ; range: [11/4,∞). The line x=2 fails the vertical line test because one input has infinitely many y-values.
A family describes a shared structure
A function family is a collection of rules with a common mathematical form, such as linear or quadratic rules.
Recognizing the family suggests a starting shape and possible behavior, but a particular function’s coefficients and domain still determine its actual graph.
Before drawing an unfamiliar formula, it helps to recognize a familiar shape inside it. These families give us starting graphs whose domains, ranges, and characteristic behavior we can predict. We will use those starting points in the next section to explain shifts and reflections.
Read change, not just height
A function is strictly increasing on an interval if whenever \(x_1<x_2\) are in that interval, \(f(x_1)<f(x_2)\). It is strictly decreasing if the output inequality is reversed. These definitions compare two inputs in the same interval; they do not say that the outputs must be positive or negative.
We have identified parent shapes and their ranges. Now read a graph from left to right to describe how its outputs change. For \(f(x)=(x-2)^2-3\), the outputs decrease on \(( -\infty,2)\) and increase on \((2,\infty)\). The vertex separates these behaviors even though many outputs near it are negative. This language helps describe transformations and, later, rates of change.
Compare two claims
A student says that \(f(x)=1/x\) decreases on its entire domain because both visible branches slope downward. Is the claim correct?
Show worked solution
It decreases separately on \((-\infty,0)\) and \((0,\infty)\). For two inputs within either interval, \(f(x_2)-f(x_1)=(x_1-x_2)/(x_1x_2)<0\). But \(-1<1\) while \(f(-1)=-1<1=f(1)\), contradicting a decrease across the entire domain. State the intervals; do not join behaviors across an excluded input.
Change the input, watch the output
Move x through the interval. Predict how y changes when x increases by one.
Translations & reflections
First compare y=x², y=x²−2, and y=(x−2)². The second moves the parabola down; the third moves it right.
Rules · How points move
| New function | Effect on the original graph |
|---|---|
| f(x)+k | Move up k units (down when k < 0). |
| f(x−h) | Move right h units (left when h < 0). |
| −f(x) | Reflect across the x-axis: (x,y) → (x,−y). |
| f(−x) | Reflect across the y-axis: (x,y) → (−x,y). |
Explore · Move and reflect a graph
Choose a parent function, then change one control at a time. Predict the result before moving the slider.
The selected reflection is applied to the parent first; the horizontal and vertical translations follow.
Common mistake · Inside shifts have the opposite sign
In f(x−2), the original input 0 now occurs at x=2, so the graph moves right. The graph of f(x+2) moves left. Reflecting x² across the y-axis leaves it unchanged; try √x to see the difference.
Try it 5 · Describe the shifts
Relative to y=3x³, describe:
- y=3(x+5)³
- y=3x³−12
- y=3(x−3)³
- y=3(x−3)³+7
Show solution
- 5 units left.
- 12 units down.
- 3 units right.
- 3 units right and 7 units up; the point (0,0) becomes (3,7).
Try it 6 · Reflect the whole rule
Let f(x)=x³−2x+1. Write the equations after reflection across each coordinate axis.
Show solution
Across the x-axis: −f(x)=−x³+2x−1. Across the y-axis: f(−x)=−x³+2x+1. Only the x-axis reflection changes the sign of the constant output term.
Once you recognize the basic shapes, you do not need to start every graph from a table of values. Think of the familiar graph as a starting picture. We will track what happens to its points when the rule shifts or reflects that picture.
Extend the point rule to scaling
For \(g(x)=a f(b(x-h))+k\), with \(b\ne0\), an original point \((u,v)\) becomes \((h+u/b,av+k)\). Solve \(b(x-h)=u\) to locate the new input; then transform the old output. This single rule combines translation, reflection, and scaling without guessing an order from the symbols.
Track an asymmetric point
Suppose \((4,3)\) lies on \(y=f(x)\). Locate its corresponding point on \(g(x)=-2f(2(x-1))+5\).
Show worked solution
Solve \(2(x-1)=4\), giving \(x=3\). The new output is \(-2(3)+5=-1\), so the point is \((3,-1)\). Multiplying the argument by \(2\) halves horizontal distances before the shift; multiplying the output by \(-2\) reflects and doubles vertical distances before adding \(5\).
Piecewise functions
Definition 4 · Different rules on different intervals
A piecewise-defined function uses different expressions on different parts of its domain.
Choose the interval containing the input before substituting. On the graph, a filled circle includes an endpoint; an open circle excludes it.
Example 4 · Choose the correct branch
f(x) =
f(−2)=−3, using the first branch.
f(−1)=0, using the middle branch.
f(2)=0, using the last branch.
Domain: ℝ. Range: (−∞,3).
The last branch gives (−∞,0]; the middle gives [0,3). The first branch adds −3, which is already included.
Try it 7 · Three branches, one function
g(x) =
Find g(−2), g(−1), g(3), then its domain and range.
Show solution
g(−2)=3, g(−1)=1, g(3)=−1. The domain is ℝ.
The first branch has range [2,∞) because its vertex at x=−3 is included. The middle branch gives (0,√2), and the last gives (−∞,0]. Together:
R=(−∞,√2) ∪ [2,∞).
Try it 8 · Sketch and state the range
g(x) =
Show solution
Draw y=3x−1 up to the filled endpoint (1,2). Draw y=x+3 from the open endpoint (1,4), extending right. Domain: ℝ; range: (−∞,2] ∪ (4,∞).
Common mistake · Range is a union of outputs
Do not read the range from the x-intervals. Find the y-values reached by each branch, then combine them. An open endpoint on one branch may still be attained elsewhere.
A single function can use different rules for different inputs. Before calculating anything, locate the input in the stated intervals and choose the matching branch. At a boundary, the endpoint symbols decide which rule applies; we do not average the two nearby outputs.
Choose a rule that matches the situation
A model is a function chosen to describe a situation. Its inputs, units, and assumptions determine which values are meaningful. A formula may accept more inputs than the situation does. Piecewise rules are useful when a price, rate, or physical behavior changes after a threshold.
Model a two-stage journey
A cyclist travels at \(12\) km/h for \(0.5\) hours, rests for \(0.25\) hours, then rides at \(8\) km/h for another \(0.5\) hours. Model total distance \(d(t)\), in kilometres, from departure. Give its domain and range, and explain why the final rule is not \(8t\).
Show worked solution
The first stage covers \(12(0.5)=6\) km. Rest adds no distance. In the final stage, \(t-0.75\) is the time spent riding since the restart; add the \(6\) km already covered. The final distance is \(10\) km, so the domain is \([0,1.25]\) hours and range \([0,10]\) km. The distance never decreases and stays constant during the rest. Using \(8t\) would wrongly apply the final speed to all elapsed time.
When you inspect a model, check its endpoint values, units, and transitions. A jump in total distance would mean instant travel; a flat segment means time passes without further distance.
Predict before you reveal
During the cyclist’s rest, several times give the same distance. Decide whether distance is still a function of elapsed time.
Intercepts & multiplicity
Definition 5 · Where the graph meets an axis
An x-intercept is a point (r,0) on the graph; solve f(r)=0 with r in the domain. The number r is a zero.
A y-intercept is (0,f(0)), provided 0 belongs to the domain.
Example 5 · Factor before graphing
g(x)=(x+3)(x−2)²(2x+3)
The zeros are −3, 2, −3/2, so the x-intercepts are (−3,0), (2,0), (−3/2,0).
g(0)=(3)(4)(3)=36, so the y-intercept is (0,36).
The factor (x−2)² makes 2 a zero of multiplicity 2.
Multiplicity · Crossing or touching
After combining identical factors, a zero r has multiplicity k if (x−r)ᵏ divides the polynomial and the remaining factor is nonzero at r.
- Odd k: the graph crosses the x-axis at r.
- Even k: the graph touches the x-axis and turns back.
- When k > 1, the graph flattens near the zero.
Try it 9 · Find all intercepts
- f(x)=(x²+3x−4)(3x+5)
- h(x)=(x³+7x²+10x)(x²−9)
Show solution
- Factor x²+3x−4=(x+4)(x−1). The x-intercepts are (−4,0), (1,0), (−5/3,0). The y-intercept is (0,−20) because f(0)=(−4)(5)=−20. All three zeros have multiplicity 1.
- Factor h(x)=x(x+5)(x+2)(x−3)(x+3). The x-intercepts are (0,0), (−5,0), (−2,0), (3,0), (−3,0). The y-intercept is (0,0). Each zero has multiplicity 1.
After learning how to move graphs, we need reliable landmarks for drawing them. Intercepts tell us where the graph meets an axis, while the multiplicity of a polynomial zero helps explain whether it crosses or touches. We will factor the rule and connect each factor with behavior near its zero.
Rational functions & holes
Definition 6 · Rational function
A rational function has the form f(x)=p(x)q(x), where p and q are polynomials and q is not the zero polynomial. Its domain excludes every real zero of the original denominator.
Example 6 · Same formula, different domains
Compare g(x)=2x²x−x² and h(x)=2x1−x.
Factoring gives g(x)=2x·xx(1−x)=2x1−x, but only for x ≠ 0,1.
Thus g has a hole at (0,0), while h includes (0,0). Both have vertical asymptote x=1. Cancellation simplifies the rule; it never restores an excluded input.
Classify a candidate x = c
- Record the zeros of the original denominator.
- Factor and cancel all common factors.
- If c is excluded and the reduced denominator is nonzero at c, there is a hole. Substitute into the reduced expression to get its height.
- If the reduced denominator is zero at c, there is a vertical asymptote.
- A zero of the reduced numerator is an x-intercept only if it belongs to the original domain.
Common mistake · 0/0 does not always mean a hole
For f(x)=x−1(x−1)²=1x−1, the original numerator and denominator both vanish at 1, but a denominator factor remains. Therefore x=1 is a vertical asymptote, not a hole.
Try it 10 · Domain and missing points
Classify all excluded inputs of r(x)=x(x−1)(x+2)x²(x+2).
Show solution
Original denominator: x²(x+2); hence D=ℝ ∖ {0,−2}. The reduced expression is (x−1)/x.
At x=0 the reduced denominator remains zero: vertical asymptote x=0. At x=−2 the reduced expression gives 3/2: hole (−2,3/2). The x-intercept is (1,0); there is no y-intercept.
We have used intercepts to locate points on a graph. A quotient brings a new issue: some inputs may not belong to the graph at all. Simplifying the expression helps us see its shape, but it does not restore an input excluded by the original denominator. Keep a record of those exclusions before canceling anything.
Asymptotes & end behavior
Definition 7 · Approaching a line
Vertical asymptote x=a: f(x) tends to +∞ or −∞ as x approaches a from at least one side.
Horizontal asymptote y=b: f(x) tends to b as x tends to +∞ or −∞.
A vertical asymptote describes behavior near a particular input; a horizontal asymptote describes behavior far to the left or right. These are limiting statements, so a few plotted points alone do not establish an asymptote.
Horizontal asymptotes · Compare degrees
| Numerator degree n vs. denominator degree m | Horizontal asymptote |
|---|---|
| n < m | y=0 |
| n = m | y = (leading coefficient of numerator)/(leading coefficient of denominator) |
| n > m | No horizontal asymptote |
Example 7 · Three different outcomes
- x+1x²+x−6: numerator degree 1 is smaller than denominator degree 2, so y=0.
- x²−4x²−6x+9: equal degrees and leading coefficients 1 and 1, so y=1.
- x³x²+1: numerator degree is larger, so no horizontal asymptote. Division gives x−x/(x²+1); the remainder tends to 0, so the slant asymptote is y=x.
Common mistake · A horizontal asymptote is not a barrier
The graph may cross a horizontal asymptote at finite x. For example, x/(x²+1) crosses its asymptote y=0 at (0,0). A rational function cannot meet a vertical asymptote because that input is excluded.
Try it 11 · Name the asymptotes
Find the vertical and horizontal asymptotes of n(x)=x²−4x²−6x+9. Describe its behavior near x=3.
Show solution
Factor the denominator: (x−3)². No factor cancels, so the vertical asymptote is x=3. Equal degrees give y=1. Near x=3, the numerator approaches 5 and the squared denominator approaches 0 through positive values. Thus n(x) tends to +∞ from both sides.
Approaching a line
An asymptote describes a limiting relationship between a graph and a line. A vertical asymptote concerns unbounded outputs near a finite input; a horizontal or slant asymptote concerns behavior far to the left or right.
It is a guide to behavior, not automatically a boundary that the graph cannot cross.
An excluded input told us where the original formula fails. We now ask a different question: what do nearby outputs do, and what happens far from the origin? Those behaviors determine asymptotes. Keep holes separate from unbounded behavior, and remember that a horizontal asymptote describes the ends of a graph, not a barrier it can never cross.
Graph a rational function
Use this five-step method. A graph should agree with the domain, intercepts, asymptotes, and interval signs simultaneously.
Example 8 · Build the graph in layers
g(x)=x²−92x²+7x+3=(x−3)(x+3)(2x+1)(x+3)
- Domain. D=ℝ ∖ {−3,−1/2}. The reduced rule is (x−3)/(2x+1).
- Intercepts. (3,0) and (0,−3). Input −3 is excluded, so it is not a zero.
- Holes and asymptotes. Hole (−3,6/5); vertical asymptote x=−1/2; horizontal asymptote y=1/2.
- Sign table. Use all zeros and excluded inputs as boundaries.
| Interval | (−∞,−3) | (−3,−1/2) | (−1/2,3) | (3,∞) |
|---|---|---|---|---|
| Sign of g | + | + | − | + |
5. Sketch. Plot the features, use the signs, and draw separate smooth branches. As x approaches −1/2 from the left, g(x) → +∞; from the right, g(x) → −∞. As x → ±∞, g(x) → 1/2.
Hole (−3,1.2) is marked with an open circle; the intercepts use filled circles.
Try it 12 · Full graphing practice
Analyze and sketch f(x)=2x+8x²−4. Find its domain, intercepts, asymptotes, and sign table before revealing the solution.
Show solution
Factor: f(x)=2(x+4)/[(x−2)(x+2)]. There is no cancellation.
- Domain: ℝ ∖ {−2,2}.
- Intercepts: (−4,0) and (0,−2).
- Vertical asymptotes: x=−2 and x=2.
- Horizontal asymptote: y=0. No holes.
| Interval | (−∞,−4) | (−4,−2) | (−2,2) | (2,∞) |
|---|---|---|---|---|
| Sign | − | + | − | + |
At −2, the left limit is +∞ and the right limit is −∞. At 2, the left limit is −∞ and the right limit is +∞.
Your graphing checklist
Use these checks while solving. They reset when you leave the lesson.
Let us bring the pieces together. Domain restrictions, intercepts, holes, and asymptotes each tell us something different about the same graph. Our task is to make those clues agree, then use a few carefully chosen points to decide how the branches connect.
Selected review problems
Try these without the notes, then compare your reasoning with the solutions. Keep your domain restrictions even when expressions simplify.
Review 1 · Evaluate and form a difference quotient
Let f(x)=2x−4 and g(x)=3x²−5x. Find:
- f(x+h)−f(x)h
- 2f(3)+g(2)
- g(y+3)
Show solution
- 2/(x+h−4)−2/(x−4)h=−2hh(x+h−4)(x−4)=−2(x+h−4)(x−4).
Restrictions: h ≠ 0, x ≠ 4, x+h ≠ 4.
- 2(−2)+(12−10)=−2.
- 3(y+3)²−5(y+3)=3y²+13y+12.
Review 2 · Domain with two restrictions
m(x)=3+√(x+6)x²−5x−24
Show solution
The radical requires x ≥ −6. The denominator factors as (x−8)(x+3), so exclude −3 and 8.
D=[−6,−3) ∪ (−3,8) ∪ (8,∞).
Review 3 · A hole can survive cancellation
Find the domain, intercepts, hole, and asymptotes of g(x)=(2x+5)(x−1)x²−1 and sketch the graph.
Show solution
g(x)=(2x+5)/(x+1)=2+3/(x+1), with original restrictions x ≠ −1,1.
- Domain: ℝ ∖ {−1,1}.
- x-intercept: (−5/2,0); y-intercept: (0,5).
- Hole: (1,7/2).
- Vertical asymptote: x=−1; horizontal asymptote: y=2.
Review 4 · Polynomial zeros
For h(x)=−(x+2)²(x−3), find the domain and intercepts, then describe what happens at each zero.
Show solution
Domain: ℝ. x-intercepts: (−2,0) and (3,0). y-intercept: (0,12). The graph touches at −2 (multiplicity 2) and crosses at 3 (multiplicity 1). Its leading term is −x³, so it rises to the left and falls to the right.
Review 5 · Complete the quadratic picture
For m(x)=2x²+5x−3, find the domain, range, intercepts, and vertex.
Show solution
Factor: (2x−1)(x+3). x-intercepts: (1/2,0) and (−3,0). y-intercept: (0,−3).
Vertex: (−5/4,−49/8). Since the parabola opens upward, domain: ℝ; range: [−49/8,∞).
Before you move on
Can you explain why a repeated output is allowed, why cancellation does not restore a missing input, and why a horizontal asymptote may be crossed? If so, you are connecting the formulas to their meaning.
Additional practice: mixed exercises
Domain practice
Find the real domain. Factor each denominator and exclude its zeros.
- \(f(x)=(x^2+2x-8)/(x^2-9)\)
- \(h(x)=(x+1)/(x^2+3x-40)\)
- \(f(x)=(x+1)/(x^2+x-6)\)
- \(n(x)=(x^2-4)/(x^2-6x+9)\)
- \(m(x)=x^3/(x^2+1)\)
Show worked solution
- \(x^2-9=(x-3)(x+3)\), so the domain is \(\mathbb R\setminus\{-3,3\}\).
- The denominator is \((x+8)(x-5)\), so the domain is \(\mathbb R\setminus\{-8,5\}\).
- The denominator is \((x+3)(x-2)\), so the domain is \(\mathbb R\setminus\{-3,2\}\).
- The denominator is \((x-3)^2\), so the domain is \(\mathbb R\setminus\{3\}\).
- \(x^2+1>0\) for every real \(x\), so the domain is \(\mathbb R\).
Reverse graphing and domain challenges
Graph features are pieces of evidence about a rule. We now work in the reverse direction: build a rule from those features, then check every restriction against the original expression. A graph sketch alone is not enough to decide whether a missing input or output is actually attained.
Multi-step · Build a rational function from its graph features
Construct one rational function with vertical asymptote \(x=2\), horizontal asymptote \(y=1\), a hole at \(x=-1\), and x-intercept \(x=3\). Find the hole height and y-intercept. Justify that your function has the stated features.
Hint
First build a ratio of two linear factors for the intercept and vertical asymptote. Then introduce a common factor for the hole.
Show worked solution
For allowed inputs, \(f(x)=(x-3)/(x-2)=1-1/(x-2)\). The uncancelled denominator produces the vertical asymptote \(x=2\); equal leading coefficients give the horizontal asymptote \(y=1\). At \(x=-1\), the reduced rule would give \(4/3\), so the hole is \((-1,4/3)\). At \(x=3\), the original denominator is nonzero, giving the x-intercept \((3,0)\). At \(x=0\), the value is \(3/2\), giving y-intercept \((0,3/2)\). The common factor must remain visible in the original rule or be recorded as a domain restriction.
Reasoning · Does a hole always remove a range value?
Compare \(f(x)=(x^2-1)/(x-1)\) and \(g(x)=(x^3-x^2)/(x-1)\), both with their original real domains. Determine their ranges and explain why deleting the input \(1\) has different effects.
Hint
Simplify without restoring the excluded input. Ask whether another allowed input produces the height of the hole.
Show worked solution
For \(x\ne1\), \(f(x)=x+1\). Only \(x=1\) would produce \(y=2\), so its range is \(\mathbb R\setminus\{2\}\). For \(x\ne1\), \(g(x)=x^2\). The missing point is \((1,1)\), but \(g(-1)=1\), so that output remains. Every \(y\ge0\) is obtained: use \(x=-\sqrt y\), which is never \(1\). Thus the range is \([0,\infty)\). A hole removes an input; it removes an output only when no other permitted input gives that output.
Advanced / Honors · Classify an entire family
For real \(a\), let \(f_a(x)=(x^2-a^2)/(x-a)\). Determine the domain, range, hole and x-intercepts for every \(a\). Explain why \(a=0\) is a special case.
Hint
Factor the numerator, preserve the original restriction, and check whether the proposed zero is permitted.
Show worked solution
The domain is \(\mathbb R\setminus\{a\}\). For allowed \(x\), factoring gives \(f_a(x)=x+a\); the graph is that line with a hole at \((a,2a)\). Solving \(y=x+a\) gives the unique input \(x=y-a\), which is excluded precisely when \(y=2a\). Hence the range is \(\mathbb R\setminus\{2a\}\). A zero would require \(x=-a\), and it is allowed exactly when \(-a\ne a\), or \(a\ne0\). Therefore there is one x-intercept \((-a,0)\) if \(a\ne0\), and none if \(a=0\). A zero of the original numerator is not sufficient: the original denominator must also be nonzero.
