If f(x) = 2x − 3, evaluate f(5) and f(−1).
Multiplying a whole function by a constant k stretches or compresses every output (y-value) by that factor. If k is negative, the graph also reflects across the x-axis.
| g(x) = 2f(x), f(3) = 4 | g(3) = 2f(3) = 2(4) = 8 — the y-value doubles, x stays the same |
| h(x) = −½f(x), f(0) = −2 | h(0) = −½(−2) = 1 — compressed AND flipped over the x-axis |
| f has zeroes at −4 and 8, y-int of −32 | g(x)=½f(x) keeps the SAME zeroes (−4, 8) but its y-int becomes −16 |
| Quadratic f(x) has turning point (8,−6). g(x)=5f(x) | g(8) = 5f(8) = 5(−6) = −30 → new turning point (8,−30) |
All transformations follow two rules: outside the function acts on outputs (vertical, moves as written); inside acts on inputs (horizontal, moves opposite). f(x) + k slides up k; f(x − h) slides right h; a negative out front flips over the x-axis; |a| > 1 stretches tall.
| Describe g(x) = −f(x) + 3. | Flip over the x-axis, then rise 3. Outside operations, in order. |
| Describe g(x) = f(x + 5) − 2. | Left 5 (inside, opposite), down 2 (outside, as written). |
If f(x) = x² and g(x) = 3f(x), describe what happens to the graph.
| g(x) = f(2x), so g(3) = f(2·3) = f(6) | To find g(3), first double the input, THEN plug into f |
| g(x) = f(½x) | Horizontal STRETCH by a factor of 2 — each x-coordinate on f's graph doubles |
| f has zeroes at −6 and 12. g(x) = f(3x) | Divide each zero by 3: g's zeroes are −2 and 4 (compression by 3) |
| f has zeroes at −8 and 12. h(x) = f(¼x) | Divide by ¼ = multiply by 4: h's zeroes are −32 and 48 (stretch by 4) |
| h(x) = 2f(3x) | Vertical stretch by 2 (y-values double) AND horizontal compression by 3 (x-values divided by 3) — do the horizontal step first when evaluating a specific point |
Multiplying the input stretches horizontally — and inversely: f(2x) squeezes the graph to half its width (things happen twice as fast), while f(x/2) stretches it to double width (twice as slow). Inside always behaves opposite to what it looks like.
| g(x) = f(3x) does what? | Compresses horizontally by a factor of 1/3 — the graph's features occur at one-third the x-distance. |
| Compare y = √(x/4) with y = √x. | Stretched horizontally by 4: y = √x hits 2 at x = 4, but √(x/4) doesn't reach 2 until x = 16. |
A movie ticket costs $12. Write an equation for total cost C as a function of tickets bought, t.
A continuous variable can take on any real number value between its extremes. A discrete variable only takes on isolated, unconnected values — usually because the input must be a whole number (you can't buy 2.5 tickets or have 3.7 people in line).
| Distance traveled at a constant speed, as a function of time | Continuous — time flows smoothly, no skipped values |
| Money raised at a bake sale, as a function of cookies sold | Discrete — the number of cookies must be a whole number |
| Which is discrete: (a) volume of an ice cube melting over time, or (b) number of pets and food purchased? | (b) is discrete — the number of pets must be a whole number; volume and time in (a) are both continuous |
| T = 40 − 3r (ride tickets remaining after r rides). Solve T = 0. | r = 13.3 — NOT viable, since the number of rides must be a whole number |
Discrete functions live only at separate points (whole-number inputs: people, tickets, terms of a sequence) — the graph is dots, not a connected curve. Sequences are discrete functions in disguise: input = term number, output = term. Recursive formulas say 'next = rule(previous)' and must state the starting term.
| Should the graph of 'cost vs number of tickets' be connected? | No — you can't buy 2.5 tickets. Plot dots at whole numbers only. |
| Recursive: a₁ = 5, an = 2an−1. List four terms. | 5, 10, 20, 40 — each term doubles the last. Without a₁, the rule has nowhere to start. |
A population starts at 200 and grows by 15 per year. Write a linear model P(t).
Given a starting value and one additional data point, you can build EITHER a linear model (constant amount added each step) or an exponential model (constant percent multiplied each step). Which one you should trust depends on how well it matches a THIRD data point.
| Tank holds 150 gal at t=0, 180 gal at t=1 | Linear: slope = (180−150)/1 = 30, so V = 30t+150. Exponential: % change = 30/150 = 20%, so V = 150(1.20)ᵗ |
| At t=10, actual volume is 500 gal | Linear predicts 30(10)+150=450. Exponential predicts 150(1.2)¹⁰≈929. Linear is much closer — it's the better model here. |
| c(n) = 6.50n + 1,245 (tire factory cost) | 6.50 = cost added per additional tire; 1,245 = fixed cost with zero tires produced |
| p(m) = 135(1.28)ᵐ (bacteria population) | 135 = starting population at m=0; 1.28 means the population grows 28% every minute |
Choosing linear vs exponential: does the story add a fixed amount per period (salary +$2,000/yr, constant differences) or multiply by a fixed factor (population ×1.03/yr, constant ratios)? Match the story's verb to the model. Exponential always wins eventually, even from behind.
| Job A: $40k + $2k raises. Job B: $40k with 4% raises. Salary in year 10? | A: 40 + 2(9) = $58k. B: 40(1.04)⁹ ≈ $56.9k — but B passes A around year 12 and never looks back. |
| A table's y-values: 100, 90, 81, 72.9 — which model? | Ratios all 0.9 → exponential decay: y = 100(0.9)x. |
Simplify: −2(3x − 5) + 4x
A step function's output stays perfectly constant over an interval, then jumps instantly to a new constant value. Graphically it's a series of horizontal segments. Pay close attention to which endpoint of each interval is closed (filled dot, ≤) vs. open (empty circle, <) — that's what makes it a true function.
| Electrician charges $40 per hour or any part of an hour | 1.2 hours still costs $80 (rounds UP to the next full hour charged) |
| f(x) = 2 for 0≤x<3; f(x) = 5 for 3≤x<5; f(x) = −4 for 5≤x≤10 | f(2.7)=2 (falls in first interval); f(5)=−4 (5 belongs to the THIRD piece, since 5≤x) |
| Parking: $3 for the 1st hour or part, +$3 each additional hour or part, $30 max | Parked for 5 hrs 22 min → falls in the "5 to 6 hour" bracket → pay for 6 full hours = 6×$3 = $18 |
Step functions model 'jump pricing': the output holds constant across an interval, then leaps at a breakpoint. The killer detail is boundary handling — does exactly 2 hours bill as tier one or tier two? Open/closed circles on the graph encode who owns each boundary point.
| Parking: $6 per hour or any part. Cost of 3 hr 5 min? | Any part of the 4th hour bills fully: 4 × $6 = $24. |
| Ceiling vs floor | 'Round up' stories (shipping, parking) use ceiling; 'complete units only' stories (full boxes packed) use floor. |
Find the slope of the line through (14,6) and (20,0).
A piecewise linear function is made of straight-line pieces that apply over different domain intervals. Each piece is a full linear equation (y = mx + b), but only for its stated x-range. Together, every x-value still produces exactly ONE y-value — that's what keeps it a valid function.
| Mateo walks 6 blocks in 9 min, then stands still 5 min, then walks home in 6 min | Piece 1 (0≤t≤9): rises from 0 to 6, slope 6/9 = 2/3, so D = (2/3)t. Piece 2 (9≤t≤14): constant, D=6. Piece 3: falls from 6 to 0. |
| Points (14,6) and (20,0) on the last segment | m = −1 (found above). Using (14,6): 6 = −1(14)+b → b = 20. So D = −t + 20 for 14≤t≤20. |
| Setting the piece −½x+4=0 (valid only for 2≤x≤6) equal to zero gives x=8 | x=8 is OUTSIDE the interval 2≤x≤6, so it is NOT a viable zero for that piece — the function has no zero there |
A piecewise function is several rules stitched together, each owning its own slice of inputs. To evaluate: first find which condition the input satisfies, then use only that rule. When graphing, mark endpoints carefully — closed dot where the piece includes the boundary, open where it doesn't.
| f(x) = x + 1 for x < 2; f(x) = 3x − 4 for x ≥ 2. Find f(2) and f(0). | f(2): second rule (2 ≥ 2) → 2. f(0): first rule → 1. |
| Why can't both pieces claim x = 2? | One input, one output — overlapping claims would break the function rule. Conditions must not overlap. |
Evaluate g(x) = x² − 4 at x = 0, and h(x) = 2x + 5 at x = 0.
Piecewise functions don't have to be built from lines only — a piece can be quadratic, square-root, or cube-root shaped. Build a table of values SEPARATELY for each piece, using only the x-values that belong to that piece's interval.
| f(x) = 2x+5 for x≤0; f(x) = ½x²−4 for x>0 | Left of the origin the graph is a straight line (constant rate of change); right of the origin it's a parabola (rate of change is NOT constant) |
| Left of origin: parabola opening down, y-int 4. Right of origin: parabola opening up, y-int −4. | This matches f(x) = 4−x² for x<0, f(x) = x²−4 for x≥0 |
| f(x)=g(x)? Compare a piecewise curve to a line g(x)=x−6 | Graph both and look for intersection points; only accept x-values where BOTH pieces are valid on their stated intervals |
Real pricing is often piecewise: a flat base up to a threshold, then a per-unit rate beyond it. The key move is charging the extra rate only on the amount past the threshold: base + rate × (x − threshold). Build each piece from its own sentence in the story.
| Electric: $20 base covers 100 kWh; $0.15 per kWh after. Bill for 260 kWh? | 20 + 0.15(260 − 100) = 20 + 24 = $44. |
| Taxi: $3 flag drop + $2/mile after the first mile. Cost of a 6-mile ride? | 3 + 2(6 − 1) = $13. |
Solve (x−2)² = 9 for x.
Height over time for a thrown or dropped object is quadratic: h(t) = −16(t−h)²+k in feet-and-seconds (or −4.9(t−h)²+k in meters). The vertex (h,k) directly gives the time and height of the maximum point.
| h(t) = −16(t−2)² + 144 | Vertex form shows max height 144 ft at t=2 sec, without any extra calculation |
| h(0) = −16(0−2)²+144 = −16(4)+144 = 80 | The projectile was launched from a height of 80 ft |
| −16(t−2)²+144 = 0 | (t−2)²=9 → t−2=±3 → t=5 or t=−1 (reject negative) → lands at t = 5 sec |
| 160 ft of fence splits a garden into 3 equal strips, width x, overall length y | Fencing equation: 4x+2y=160 → y=−2x+80. Area A=xy=x(−2x+80)=−2x²+80x |
| Maximizing A = −2x²+80x | Vertex occurs halfway between the zeroes (x=0 and x=40), so x=20 gives max area of 800 sq ft |
Quadratic modeling headlines: the vertex is the max/min (peak height, best price), the roots are where the quantity hits zero (landing time, break-even), and the y-intercept is the starting value. Translate the question into which landmark it wants, then use −b/2a, factoring, or the formula accordingly.
| R(p) = −5p² + 100p. What price maximizes revenue? | Vertex: p = 100/10 = $10, giving R = $500. |
| h(t) = −16t² + 32t + 48. When does it land? | −16(t² − 2t − 3) = −16(t − 3)(t + 1) = 0 → t = 3 s (reject −1). |
Round 43.86 to the nearest tenth.
No real-world measurement is perfectly exact — every number you measure was rounded to some level of precision. A calculation can never be MORE precise than its least precise input, even if the math itself produces more decimal places.
| Length = 2.7 m, width = 1.4 m (both to the nearest tenth) | Area = 2.7×1.4 = 3.78, but must be rounded to 3.8 m² — matching the tenths precision of the inputs |
| h(t)=−4.9t²+24t+3, evaluated at t=1.7 (rounded to nearest tenth) | h(1.7)=29.639 meters exactly, but since t was only known to the tenth, report h(1.7) ≈ 29.6 meters |
| V(t)=362−12.8t, initial 362 gal known to the nearest whole gallon | V(7)=272.4, but round to the least precise input's level: 272 gallons (whole number) |
| Newborn weights measured to the nearest tenth of a pound | Mean and standard deviation should ALSO be reported to the nearest tenth — not carried out to more decimal places just because the calculator shows them |
Models are maps, not territory: they're built from limited data and quiet assumptions. Check the practical domain (inputs that make real-world sense), beware extrapolation, and round answers to sensible precision — predicting 463.2891 bacteria pretends to more accuracy than the model owns.
| A height model gives h(−1) = 20 ft. Meaningful? | No — t = −1 is before launch. Negative time sits outside the practical domain even though the algebra computes fine. |
| Population model predicts 12,268.437 people. | Report ≈ 12,300 or 12,268 — fractional people signal false precision, not better math. |
By hand: if f(x) = 2x−3, find f(5).
This is the calculator authorized for this course. Unit 11 pulls together everything from the year — transformations, piecewise/step functions, and quadratic models — so this lesson is your one-stop toolkit for checking any of them.
The calculator settles modeling questions numerically: Maximum/Minimum finds the vertex, Zero finds landing/break-even points, Intersection compares two models. Set a window that fits the story first (time ≥ 0, sensible height range) — a good window is half the work.
| Find when two savings plans are equal. | Graph both, Menu → Analyze → Intersection. The x-coordinate is the break-even time. |
| Maximum returns a slightly-off decimal like 1.9999998. | That's numerical rounding — the true answer is 2. Confirm with −b/2a by hand. |
These problems pull from earlier units on purpose. Switching between skills feels harder in the moment, but it helps you remember longer and matches how a real exam mixes topics. Try each one on paper first, then reveal.Estos problemas provienen de unidades anteriores a propósito. Cambiar de una destreza a otra se siente más difícil en el momento, pero te ayuda a recordar por más tiempo y se parece a cómo un examen real mezcla los temas. Intenta cada uno en papel primero, luego revela la respuesta.