Sec 5 SN · Real Functions · Deep Study
Real
Functions
Piecewise, inverse, absolute value, and rational functions all follow logical patterns. This guide builds each from first principles with diagrams, worked examples, and checkpoints to verify your understanding.
📐 7 sections
🔍 19 worked examples
✅ 13 checkpoint questions
💡 Intuition-first
1
Piecewise Functions
Different rules for different parts of the domain
A piecewise function is defined by different formulas for different intervals of x. The function is still a function — each x gives exactly one y — but the rule changes depending on where x sits.
Think of a taxi fare: there might be a flat fee for the first 2 km, then a per-km rate after that. Two different rules, one function.
📐
Standard notation:
f(x) = { rule₁ if condition₁
{ rule₂ if condition₂
{ rule₃ if condition₃
The conditions must cover all required x values and must not overlap.
f(x) = { rule₁ if condition₁
{ rule₂ if condition₂
{ rule₃ if condition₃
The conditions must cover all required x values and must not overlap.
Open vs Closed Endpoints
| Symbol | Meaning | Inequality | Dot style on graph |
|---|---|---|---|
| Closed bracket [ | Endpoint is included | ≤ or ≥ | ● filled dot |
| Open bracket ( | Endpoint is excluded | < or > | ○ open dot |
At a boundary point, only ONE piece can own that x value. If the boundary x = 1 uses ≥ in piece 2, then piece 2 has the filled dot at x = 1 and piece 1 must use < (open dot).
Evaluating a Piecewise Function
To find f(a): first determine which condition a satisfies, then apply only that piece's formula. Never mix pieces.
★ Basic
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Evaluating a piecewise at several points
Let f(x) = { x² if x < 0 ; 3 if x = 0 ; 2x − 1 if x > 0 }. Find f(−3), f(0), f(4).
1
f(−3): check which piece applies
x = −3 satisfies x < 0 → use piece 1: f(x) = x²
f(−3) = (−3)² = 9
2
f(0): check which piece applies
x = 0 satisfies x = 0 → use piece 2: f(x) = 3
f(0) = 3
3
f(4): check which piece applies
x = 4 satisfies x > 0 → use piece 3: f(x) = 2x − 1
f(4) = 2(4) − 1 = 7
Answer: f(−3) = 9, f(0) = 3, f(4) = 7
★★ Medium
Show solution
Testing for continuity at a boundary
Is f(x) = { 2x + 1 if x < 2 ; x² − 1 if x ≥ 2 } continuous at x = 2?
1
Value from the left (approaching x = 2 from below)
Use piece 1 with x → 2:
2(2) + 1 = 5
2
Value at x = 2 (the actual function value)
x = 2 satisfies x ≥ 2 → use piece 2:
f(2) = (2)² − 1 = 3
3
Compare
Left-hand value = 5, actual value = 3. They differ.
A jump discontinuity exists at x = 2. The graph has a gap — the purple piece ends at (2, 5) (open) and the orange piece starts at (2, 3) (closed).
Answer: Not continuous at x = 2 — there is a jump discontinuity
★★★ Hard
Show solution
Finding a constant to make a piecewise continuous
Find the value of k that makes f(x) = { kx + 3 if x < 1 ; x² + k if x ≥ 1 } continuous at x = 1.
1
Set up the continuity condition
For continuity, the left-hand value must equal the right-hand value at x = 1.
Left: k(1) + 3 = k + 3
Right: (1)² + k = 1 + k
2
Solve for k
k + 3 = 1 + k
3 = 1 ← this is always false!
No value of k makes this continuous. The two pieces always differ by 2 at x = 1 regardless of k. The function has an unavoidable jump of 2 at x = 1.
Answer: No value of k makes f continuous at x = 1 — the jump is always 2
✅ Checkpoint 1A
A piecewise function is defined as f(x) = { 3 if x ≤ −1 ; x² if x > −1 }. Is the function continuous at x = −1?
From the left: f(−1) = 3 (closed — included). From the right: approaching x = −1 gives (−1)² = 1. Since 3 ≠ 1, there is a jump at x = −1 — the function is discontinuous there.
✅ Checkpoint 1B
Given f(x) = { −2x if x < 0 ; x + 5 if x ≥ 0 }, find f(−4), f(0), f(7).
f(−4) = −2(−4) = 8
f(0) = 0 + 5 = 5 (x=0 satisfies x ≥ 0)
f(7) = 7 + 5 = 12
2
Graphing Piecewise Functions
A reliable step-by-step process
- Step 1: Identify each piece, its formula, and its interval.
- Step 2: Build a table of values for each piece — include at least the boundary points and one interior point.
- Step 3: Check whether each boundary point is open or closed. Mark the dot accordingly.
- Step 4: Graph each piece separately — different colour or style helps.
- Step 5: Do NOT connect pieces across a gap. Leave a visible jump if the function is discontinuous there.
💡
After graphing, use the vertical line test to confirm it's still a function. Each vertical line should cross the graph at most once. Piecewise functions always pass this test if the conditions don't overlap.
★★ Medium
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Graphing a three-piece function
Graph f(x) = { −1 if x < −2 ; x + 1 if −2 ≤ x ≤ 2 ; 3 if x > 2 }.
1
Piece 1: constant f(x) = −1 for x < −2
Horizontal line at y = −1. Open circle at x = −2 (because < not ≤).
2
Piece 2: linear f(x) = x + 1 for −2 ≤ x ≤ 2
At x = −2: y = −1 (closed). At x = 2: y = 3 (closed). Draw the segment.
3
Piece 3: constant f(x) = 3 for x > 2
Horizontal line at y = 3. Open circle at x = 2.
Pieces 1 and 2 agree at x = −2 (both give y = −1), so it is continuous there. Pieces 2 and 3 agree at x = 2 (both give y = 3), so it is also continuous at x = 2.
Answer: Continuous everywhere — all pieces connect smoothly
★★ Medium
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Domain and range of a piecewise
Find the domain and range of f(x) = { −x if x < 0 ; x² if 0 ≤ x ≤ 3 }.
1
Domain
The function is defined for x < 0 (piece 1) and for 0 ≤ x ≤ 3 (piece 2).
Domain = (−∞, 0) ∪ [0, 3] = (−∞, 3]
2
Range of piece 1: f(x) = −x for x < 0
As x approaches 0 from the left, f(x) = −x approaches 0 (but doesn't include it since x < 0 means f(x) > 0). As x → −∞, −x → +∞.
Piece 1 range: (0, +∞)
3
Range of piece 2: f(x) = x² for 0 ≤ x ≤ 3
At x = 0: f = 0. At x = 3: f = 9. Since x² is increasing on [0, 3]:
Piece 2 range: [0, 9]
4
Combine ranges
Total range = [0, 9] ∪ (0, +∞) = [0, +∞)
Answer: Domain: (−∞, 3] | Range: [0, +∞)
★★★ Hard
Show solution
Writing a piecewise rule from a description
A parking garage charges $5 flat for the first hour, then $2 per additional hour (or part thereof), up to a maximum of $15. Write f(t) as a piecewise function for t ∈ (0, 6].
1
Identify the break points
The rule changes at t = 1 (flat fee ends) and at the point where cost hits $15.
2
Find where the cap kicks in
Cost = 5 + 2(t − 1) = 15
2(t − 1) = 10 → t − 1 = 5 → t = 6
At t = 6 the cost reaches $15, so the cap applies exactly at t = 6.
3
Write the piecewise rule
f(t) = { 5 if 0 < t ≤ 1
5 + 2(t−1) if 1 < t < 6
15 if t = 6 }
Answer: f(t) = 5 for 0 < t ≤ 1; 5+2(t−1) for 1 < t ≤ 5.5; 15 for t = 6
✅ Checkpoint 2
For f(x) = { 4 if x < −1 ; x + 5 if −1 ≤ x ≤ 0 ; −x + 5 if x > 0 }, is f continuous at x = −1? At x = 0?
At x = −1: Left-hand value = 4; piece 2 value = (−1) + 5 = 4. They match → continuous.
At x = 0: Piece 2 value = 0 + 5 = 5; piece 3 value (as x → 0⁺) = −0 + 5 = 5. They match → continuous.
The function is continuous everywhere.
3
What is an Inverse Function?
The idea before the mechanics
The inverse of a function f, written f⁻¹, undoes f. If f takes x to y, then f⁻¹ takes y back to x.
💭
Think of it like a machine with a reverse switch:
f is "double and add 1" → f(3) = 7. The inverse is "undo that" → f⁻¹(7) = 3, by subtracting 1 and halving. The output of f becomes the input of f⁻¹.
f is "double and add 1" → f(3) = 7. The inverse is "undo that" → f⁻¹(7) = 3, by subtracting 1 and halving. The output of f becomes the input of f⁻¹.
When Does the Inverse Exist as a Function?
A function has an inverse that is also a function if and only if f is one-to-one — meaning every y value comes from at most one x value.
| Test | What it checks | Result if it passes |
|---|---|---|
| Horizontal line test | Draw any horizontal line — does it cross the graph more than once? | If no line crosses twice: one-to-one → inverse is a function |
| Algebraic check | Can two different x-values give the same y? Set f(a) = f(b) and check if a = b must follow. | If a = b is forced: one-to-one |
⚠️
Quadratic functions like y = x² are NOT one-to-one (e.g. f(2) = f(−2) = 4). Their inverse relation is a sideways parabola — only a function after restricting the domain (e.g. x ≥ 0). Always check this before finding the inverse.
Domain of f
= Range of f⁻¹
Range of f
= Domain of f⁻¹
Verify identity
f(f⁻¹(x)) = x
Graphical test
Reflect over y = x
★ Basic
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Verifying two functions are inverses
Show that f(x) = 4x − 2 and g(x) = (x + 2)/4 are inverses of each other.
1
Compute f(g(x))
f(g(x)) = 4 · (x+2)/4 − 2
= (x + 2) − 2
= x ✓
2
Compute g(f(x))
g(f(x)) = (4x − 2 + 2)/4
= 4x/4
= x ✓
3
Conclusion
Both compositions give x, so g = f⁻¹ (and f = g⁻¹).
Answer: Yes — f(g(x)) = g(f(x)) = x confirms they are inverses
★★ Medium
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Reading domain and range of f and f⁻¹
f has domain [−2, 5] and range [0, 8]. State the domain and range of f⁻¹.
1
Apply the swap rule
The domain and range of f⁻¹ are the range and domain of f, respectively — they swap.
Domain of f⁻¹ = Range of f = [0, 8]
Range of f⁻¹ = Domain of f = [−2, 5]
Answer: Domain of f⁻¹ = [0, 8]; Range of f⁻¹ = [−2, 5]
✅ Checkpoint 3
Decide whether each function is one-to-one (and thus has an inverse function): (a) f(x) = 3x + 7 (b) g(x) = x² − 4 (c) h(x) = x³
(a) f(x) = 3x + 7 — linear, strictly increasing → one-to-one ✓
(b) g(x) = x² − 4 — parabola, g(2) = g(−2) = 0 → not one-to-one ✗ (unless domain is restricted)
(c) h(x) = x³ — strictly increasing for all real x → one-to-one ✓
4
Finding the Inverse — Step by Step
The mechanical process with multiple function types
The four-step algorithm
Step 0 → Replace f(x) with y
Step 1 → Swap x and y
Step 2 → Solve for y
Step 3 → Write as f⁻¹(x) = ... and note any domain restriction
💡
Step 1 (swap) is the key insight. It geometrically corresponds to reflecting the graph over the line y = x. Everything else is just algebra.
★ Basic
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Inverse of a linear function
Find f⁻¹(x) for f(x) = 3x − 6.
1
Replace and swap
y = 3x − 6
→ swap → x = 3y − 6
2
Solve for y
x + 6 = 3y
y = (x + 6)/3
Answer: f⁻¹(x) = (x + 6)/3
★★ Medium
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Inverse of a square root function
Find f⁻¹(x) for f(x) = √(x − 2) + 3.
1
Swap
x = √(y − 2) + 3
2
Isolate the radical
x − 3 = √(y − 2)
3
Square both sides
(x − 3)² = y − 2
y = (x − 3)² + 2
Domain restriction: since √ outputs ≥ 0, we need x − 3 ≥ 0, i.e. x ≥ 3. So domain of f⁻¹ = [3, +∞).
Answer: f⁻¹(x) = (x − 3)² + 2, x ∈ [3, +∞)
★★ Medium
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Inverse of a rational function
Find f⁻¹(x) for f(x) = (2x + 1)/(x − 3).
1
Swap x and y
x = (2y + 1)/(y − 3)
2
Multiply both sides by (y − 3)
x(y − 3) = 2y + 1
xy − 3x = 2y + 1
3
Collect y terms on one side
xy − 2y = 3x + 1
y(x − 2) = 3x + 1
4
Divide by (x − 2)
y = (3x + 1)/(x − 2)
Domain restriction: x ≠ 2 (since x − 2 cannot be zero).
Answer: f⁻¹(x) = (3x + 1)/(x − 2), x ≠ 2
★★★ Hard
Show solution
Inverse with a restricted domain
f(x) = (x − 1)² + 4 with domain x ≥ 1. Find f⁻¹(x) and state its domain and range.
1
Note why a restriction is needed
Without x ≥ 1, f(x) = (x−1)² + 4 is a parabola — not one-to-one. The restriction keeps only the right half (increasing part).
2
Swap and solve
x = (y − 1)² + 4
x − 4 = (y − 1)²
√(x − 4) = y − 1 [take positive root since y ≥ 1, so y−1 ≥ 0]
y = √(x − 4) + 1
3
Find domain and range of f⁻¹
Domain of f⁻¹ = Range of f. Since f has domain x ≥ 1 and minimum value at x = 1 of f(1) = 4, the range of f is [4, +∞).
Domain of f⁻¹ = [4, +∞)
Range of f⁻¹ = [1, +∞) (= domain of f)
Answer: f⁻¹(x) = √(x − 4) + 1, domain [4, +∞), range [1, +∞)
✅ Checkpoint 4A
Find the inverse of f(x) = 2x + 4. Then verify that f(f⁻¹(x)) = x.
Swap: x = 2y + 4 → x − 4 = 2y → y = (x−4)/2
f⁻¹(x) = (x−4)/2
Verify: f(f⁻¹(x)) = 2·(x−4)/2 + 4 = (x−4) + 4 = x ✓
✅ Checkpoint 4B
Find f⁻¹(x) for f(x) = 5/(x + 2), and state where it is undefined.
Swap: x = 5/(y + 2)
x(y + 2) = 5
y + 2 = 5/x
y = 5/x − 2 (i.e. (5 − 2x)/x)
f⁻¹(x) = 5/x − 2, undefined at x = 0
5
Absolute Value Functions
Understanding the V-shape
The absolute value of a number is its distance from zero on the number line — always non-negative.
📐
Definition:
|x| = x if x ≥ 0
|x| = −x if x < 0
Geometrically: it "folds" the left half of the number line onto the right half, making all values positive.
|x| = x if x ≥ 0
|x| = −x if x < 0
Geometrically: it "folds" the left half of the number line onto the right half, making all values positive.
This folding creates the characteristic V-shape. The standard form is:
Absolute value function
y = a|x − h| + k
Vertex (h, k) · |a| = slope of each arm · a > 0 opens up, a < 0 opens down
| Feature | Value / Rule |
|---|---|
| Vertex (turning point) | (h, k) |
| Axis of symmetry | x = h |
| Opening direction | Up if a > 0 · Down if a < 0 |
| Slope of right arm | +|a| |
| Slope of left arm | −|a| (mirror image) |
| Domain | All real numbers ℝ |
| Range (a > 0) | y ∈ [k, +∞) |
| Range (a < 0) | y ∈ (−∞, k] |
Intercepts of y = a|x − h| + k
- y-intercept: set x = 0 → y = a|0 − h| + k = a|h| + k
- x-intercept(s): set y = 0 → a|x − h| = −k → |x − h| = −k/a. Then solve as an absolute value equation (may have 0, 1, or 2 solutions depending on the sign of −k/a).
★ Basic
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Reading parameters and describing the graph
For f(x) = −2|x − 3| + 8, state the vertex, opening, domain, range, and slope of each arm.
1
Read parameters directly
a = −2, h = 3, k = 8
2
Vertex, opening, slopes
Vertex: (3, 8)
Opening: DOWN (a = −2 < 0)
Right arm slope: +|−2| = +2 (going right from vertex, y rises at rate 2)
Wait — since a < 0 and graph opens down:
Right arm slope: a = −2 (decreasing to the right)
Left arm slope: +2 (decreasing to the left)
When a < 0, both arms point downward from the vertex. The right arm has slope a = −2 and the left arm has slope −a = +2. Each arm decreases as you move away from the vertex.
3
Domain and range
Domain: all reals ℝ
Range: y ≤ 8 → y ∈ (−∞, 8]
Answer: Vertex (3,8), opens down, range (−∞, 8], slopes ±2 from vertex
★★ Medium
Show solution
Finding intercepts of an absolute value function
Find all intercepts of f(x) = 3|x + 2| − 6.
1
y-intercept: set x = 0
f(0) = 3|0 + 2| − 6 = 3(2) − 6 = 0
y-intercept is (0, 0) — the graph also crosses the x-axis here!
2
x-intercept(s): set y = 0
3|x + 2| − 6 = 0
3|x + 2| = 6
|x + 2| = 2
3
Split into two cases
Case 1: x + 2 = 2 → x = 0
Case 2: x + 2 = −2 → x = −4
Answer: x-intercepts at x = 0 and x = −4; y-intercept at (0, 0)
★★★ Hard
Show solution
Writing the rule from a described graph
An absolute value function has vertex (−1, 5), passes through (2, 2), and opens downward. Find f(x).
1
Write the general form with known h, k
f(x) = a|x − (−1)| + 5 = a|x + 1| + 5
2
Use the point (2, 2) to find a
2 = a|2 + 1| + 5
2 = a(3) + 5
−3 = 3a
a = −1
3
Verify direction
a = −1 < 0 → opens downward. ✓ Consistent with the description.
Answer: f(x) = −|x + 1| + 5
✅ Checkpoint 5
For f(x) = 2|x − 4| − 6: (a) Find the vertex. (b) Find the x-intercepts. (c) Find the y-intercept. (d) State the range.
(a) Vertex: (4, −6) [h=4, k=−6]
(b) Set 2|x−4| − 6 = 0 → |x−4| = 3 → x−4 = ±3 → x = 7 or x = 1
(c) f(0) = 2|0−4| − 6 = 2(4) − 6 = 2 → y-intercept: (0, 2)
(d) a = 2 > 0 → opens up → range: [−6, +∞)
6
Absolute Value Equations and Inequalities
Splitting into cases
💭
Why split into two cases?
|something| = c means the "something" is at distance c from zero — it could be c or −c. There are exactly two positions at distance c from zero (unless c = 0 or c < 0).
|something| = c means the "something" is at distance c from zero — it could be c or −c. There are exactly two positions at distance c from zero (unless c = 0 or c < 0).
|expr| = c, c > 0
expr = c OR expr = −c
|expr| = 0
One solution: expr = 0
|expr| = c, c < 0
No solution (abs ≥ 0)
|expr| < c (c > 0)
−c < expr < c (AND)
|expr| ≤ c
−c ≤ expr ≤ c
|expr| > c (c > 0)
expr > c OR expr < −c
⚠️
Always isolate the absolute value expression before splitting into cases. If the equation is 2|x − 1| + 3 = 9, first simplify to |x − 1| = 3, then split. Splitting too early with extra terms causes errors.
★ Basic
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Solving an absolute value equation
Solve |3x − 1| = 26.
1
Split into two cases
Case 1: 3x − 1 = 26
Case 2: 3x − 1 = −26
2
Solve each case
Case 1: 3x = 27 → x = 9
Case 2: 3x = −25 → x = −25/3
Answer: x = 9 or x = −25/3
★★ Medium
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Solving an absolute value inequality (<)
Solve |2x + 1| ≤ 7.
1
Set up compound inequality
−7 ≤ 2x + 1 ≤ 7
2
Subtract 1 throughout
−8 ≤ 2x ≤ 6
3
Divide by 2
−4 ≤ x ≤ 3
Answer: x ∈ [−4, 3]
★★ Medium
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Solving an absolute value inequality (>)
Solve |4 − 3x| > 10.
1
Split into two separate inequalities (OR)
4 − 3x > 10 OR 4 − 3x < −10
2
Solve Case 1: 4 − 3x > 10
−3x > 6
x < −2 (inequality flips when dividing by −3)
3
Solve Case 2: 4 − 3x < −10
−3x < −14
x > 14/3
Answer: x < −2 or x > 14/3
★★★ Hard
Show solution
Equation with absolute values on both sides
Solve |2x − 3| = |x + 1|.
1
When two absolute values are equal, either the insides are equal or they are negatives of each other
Case 1: 2x − 3 = (x + 1)
Case 2: 2x − 3 = −(x + 1)
2
Solve Case 1
2x − 3 = x + 1
x = 4
3
Solve Case 2
2x − 3 = −x − 1
3x = 2
x = 2/3
4
Verify both solutions
x = 4: |2(4)−3| = |5| = 5; |4+1| = 5 ✓
x = 2/3: |2(2/3)−3| = |−5/3| = 5/3; |2/3+1| = 5/3 ✓
Answer: x = 4 or x = 2/3
✅ Checkpoint 6A
Solve: |5x − 2| = 13
Case 1: 5x − 2 = 13 → 5x = 15 → x = 3
Case 2: 5x − 2 = −13 → 5x = −11 → x = −11/5
✅ Checkpoint 6B
Solve: |3x + 6| > 9. Express the solution in interval notation.
Split (OR): 3x + 6 > 9 OR 3x + 6 < −9
Case 1: 3x > 3 → x > 1
Case 2: 3x < −15 → x < −5
Solution: (−∞, −5) ∪ (1, +∞)
7
Rational Functions — Overview
The hyperbola family
A rational function has the form f(x) = P(x)/Q(x) where P and Q are polynomials. The Sec 5 focus is on the simplified form that produces a hyperbola:
Rational function — Sec 5 form
f(x) = a / (b(x − h)) + k
Parameters a, b, h, k · vertical asymptote at x = h · horizontal asymptote at y = k
Vertical asymptote
x = h (denominator = 0)
Horizontal asymptote
y = k (as x → ±∞)
Domain
All reals except x = h
Range
All reals except y = k
The Role of Each Parameter
| Parameter | Effect |
|---|---|
| h | Shifts the vertical asymptote to x = h (left/right shift) |
| k | Shifts the horizontal asymptote to y = k (up/down shift) |
| a/b > 0 | Branches in quadrants 1 and 3 (relative to the asymptotes) |
| a/b < 0 | Branches in quadrants 2 and 4 (relative to the asymptotes) |
| |a/b| large | Branches are stretched farther from the asymptotes |
📐
The graph has two branches — one on each side of the vertical asymptote. Both branches approach both asymptotes but never cross them. The asymptotes are invisible guidelines, not part of the graph itself.
Intercepts of f(x) = a/(b(x − h)) + k
- y-intercept: set x = 0 → y = a/(b(0 − h)) + k = a/(−bh) + k (provided h ≠ 0)
- x-intercept: set y = 0 → a/(b(x−h)) = −k → a = −kb(x−h) → x = h − a/(kb) (provided k ≠ 0)
★★ Medium
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Analyzing a rational function
Analyze f(x) = 4/(x − 3) + 2: asymptotes, domain, range, and intercepts.
1
Identify parameters
a = 4, b = 1, h = 3, k = 2
2
Asymptotes, domain, range
Vertical asymptote: x = 3
Horizontal asymptote: y = 2
Domain: all reals except x = 3
Range: all reals except y = 2
3
y-intercept (set x = 0)
f(0) = 4/(0 − 3) + 2 = −4/3 + 2 = 2/3
4
x-intercept (set y = 0)
0 = 4/(x − 3) + 2
−2 = 4/(x − 3)
x − 3 = 4/(−2) = −2
x = 1
Answer: V.A. x=3, H.A. y=2 · y-int (0, 2/3) · x-int (1, 0)
★★ Medium
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Effect of a negative a/b ratio
For f(x) = −6/(2(x + 1)) + 3, identify all features and determine in which quadrants (relative to asymptotes) the branches lie.
1
Parameters
a = −6, b = 2, h = −1, k = 3
a/b = −6/2 = −3 < 0
2
Asymptotes and key features
Vertical asymptote: x = −1
Horizontal asymptote: y = 3
Domain: x ≠ −1; Range: y ≠ 3
3
Branch locations
Since a/b < 0, the branches lie in the 2nd and 4th quadrants relative to the asymptote intersection point (−1, 3): upper-left and lower-right.
4
Intercepts
y-int (x=0): f(0) = −6/(2·1) + 3 = −3 + 3 = 0 → (0, 0)
x-int (y=0): 0 = −6/(2(x+1)) + 3 → 6/(2(x+1)) = 3 → 2(x+1) = 2 → x = 0 → (0, 0)
The function passes through the origin (0, 0) — both the x and y intercept coincide there.
Answer: V.A. x=−1, H.A. y=3, branches in Q2 & Q4 rel. to center, passes through origin
★★★ Hard
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Finding the rule from a graph description
A hyperbola has vertical asymptote x = 2, horizontal asymptote y = −1, and passes through (4, 3). Find the rule f(x) = a/(x − h) + k.
1
Read asymptotes for h and k
h = 2, k = −1
So f(x) = a/(x − 2) − 1
2
Use the point (4, 3) to find a
3 = a/(4 − 2) − 1
3 = a/2 − 1
4 = a/2
a = 8
3
Write the rule
f(x) = 8/(x − 2) − 1
Answer: f(x) = 8/(x − 2) − 1
✅ Checkpoint 7
For f(x) = −3/(x + 4) + 1: (a) State the vertical and horizontal asymptotes. (b) Find the domain and range. (c) Find the y-intercept.
(a) Vertical asymptote: x = −4; Horizontal asymptote: y = 1
(b) Domain: all reals except x = −4; Range: all reals except y = 1
(c) f(0) = −3/(0+4) + 1 = −3/4 + 1 = 1/4 → y-intercept: (0, 1/4)