Modeling functions

1. Constant rate of change — each unit increase in x produces the same change in y. 2. The graph is a straight line.

When the data shows a constant rate of change — equal steps in x produce equal steps in y. A scatter plot that looks like a straight line suggests a linear model.

5n: cost increases by 5 per unit produced (variable cost, the gradient). 200: fixed cost regardless of production level (the y-intercept).

Yes — constant speed means equal distance in equal time intervals. Distance = 80t is linear with gradient 80.

1. Calculate gradient: m = (y₂ − y₁)/(x₂ − x₁). 2. Use y = mx + c with one point to find c. 3. Write the model.

T = −2.5(12) + 80 = −30 + 80 = 50.

m = (20 − 60)/8 = −5. Using (0, 60): T = −5t + 60.

The full equation in y = mx + c form, with numerical values for m and c, using the variables named in the context.

The population increases by 4.5 people per year.

The initial weight is 50 kg — the weight at the start (d = 0), before any distance has been covered.

State: the numerical value, the units, and what it means for the specific context. E.g. "The water level rises by 3 cm per hour."

The quantity is decreasing at a constant rate as the input variable increases.

The model gives reliable, meaningful predictions for x-values within the range of the original data (interpolation). Outside this range, the model may break down.

Check if x = 50 is within the data range. If yes: "Yes — x = 50 is within the data range so the estimate is reliable (interpolation)." If no: "Less reliable — x = 50 is outside the data range (extrapolation)."

Physically extreme or impossible values signal model breakdown — this is extrapolation far beyond the data range. Real temperatures may not follow this pattern at t = 100.

Interpolation: predicting within the data range — generally reliable. Extrapolation: predicting outside the range — less reliable, the pattern may not continue.

A parabola — a symmetric U-shape. Opens upward (∪) if a > 0, downward (∩) if a < 0.

A ball thrown upward: h(t) = −5t² + 20t + 3. Height rises, reaches a maximum, then falls — the parabolic path of projectile motion.

Linear: constant rate of change, straight line. Quadratic: changing rate of change, has a maximum or minimum turning point (vertex), curved graph.

Revenue increases, reaches a maximum at the vertex (optimal price), then decreases. There is one best price for maximum revenue.

x = −b/(2a). The y-coordinate is found by substituting this x back into the equation.

x = −(−8)/(2·2) = 2. y = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5. Vertex at (2, −5).

x = −(−6)/(2·1) = 3. f(3) = 9 − 18 + 11 = 2. Minimum value is 2 (at x = 3).

If a > 0 (parabola opens up), the vertex is a minimum. If a < 0 (parabola opens down), the vertex is a maximum.

x = 3 is the x-coordinate of the vertex, not the maximum value. The maximum value is f(3) = −9 + 18 − 5 = 4.

The formula is x = −b/(2a). Forgetting the negative gives the wrong x-value and hence the wrong vertex.

No — if a > 0 the parabola opens upward and only has a minimum. Only quadratics with a < 0 have a maximum.

h(t) = 0. Set −5t² + 20t = 0 → −5t(t − 4) = 0 → t = 0 or t = 4. Ball is on the ground at t = 0 and t = 4.

t = −20/(2·−5) = 2. h(2) = −5(4) + 40 + 1 = 21. Maximum height = 21.

n = −10/(2·−1) = 5. Maximum profit at n = 5 units.

Find x-intercepts (set P = 0, solve). P is positive between the roots if a < 0, or outside them if a > 0.

R = 0 at p = 0 and p = 40. These are the prices at which revenue is zero: free (no payment) or so expensive no one buys.

y = a · bˣ. a = initial value (y-intercept at x = 0). b = growth/decay factor per unit of x.

500 = initial value (at x = 0). 1.06 = growth factor — 6% growth per unit of x.

Growth — the output increases as x increases. The greater b is above 1, the faster the growth.

Decay — the output decreases as x increases. The closer b is to 0, the faster the decay.

P(t) = 4000 · 1.05ᵗ. Initial value a = 4000, growth factor b = 1 + 0.05 = 1.05.

Q(t) = 200 · 0.5ᵗ. Initial value a = 200, decay factor b = 0.5.

Substitute both points to get two equations. Divide one by the other to eliminate a and solve for b. Then substitute b back to find a.

P = 3000 · 1.04⁵ = 3000 · 1.2167 ≈ 3650.

y = 5 · 1.03 · x is linear, not exponential. In an exponential model, x must be the exponent: y = 5 · 1.03ˣ.

b is the growth factor, not the rate. b = 1 + rate = 1 + 0.08 = 1.08. Using b = 8 would give wildly wrong values.

No — a · bˣ is always positive when a > 0 and b > 0. A negative result always means a calculation error.

Check the ratio of successive y-values: if y₂/y₁ is approximately constant, the data is exponential.

y = 0. As x → −∞, 2ˣ → 0, so the whole expression approaches 0 from above. The x-axis is the asymptote.

P → 0. The substance/quantity decays toward zero but never fully disappears (according to the model).

y = 0. Write as a full equation. The growth model approaches 0 as x → −∞.

The model predicts the quantity approaches zero but never reaches it. In reality, the quantity may reach zero (e.g. a substance fully decays). The model is a mathematical idealisation.

y = axⁿ, where a is a constant and n is any real-number power.

Area of circle: A = πr² (power 2). Distance under gravity: s = 5t² (power 2). Surface area ∝ length² for similar shapes.

Power model: x is the base, n is a fixed exponent. Exponential: x is the exponent, b is a fixed base. Very different shapes for large x.

y increases by a factor of 2² = 4. Power models scale multiplicatively: doubling x multiplies y by 2ⁿ.

y = 2x³ is a power model — x is the base. y = 2 · 3ˣ is exponential — x is the exponent.

Exponential always eventually grows faster than any power model. Even x¹⁰⁰ is eventually overtaken by 2ˣ.

No — exponential y = a · bˣ passes through (0, a), not the origin (unless a = 0). A power model with n > 0 passes through (0, 0).

Exponential — x is in the exponent. Base 0.7 means decay. It is NOT a power model.

Power regression (PwrReg on TI-84). Returns a and b for y = axᵇ.

y = 3.25x^0.812 (all values to 3 s.f.).

When the scatter plot shows a curved relationship (not straight), the data passes near the origin, and a straight line clearly doesn't fit the pattern.

Very strong fit — 97% of variation in y is explained by the power model. It is a very good fit for the data.

Mass grows slightly faster than the square of diameter. Doubling d multiplies M by 2^2.1 ≈ 4.3.

The model was built from data in a limited range. Using it for x well beyond that range is extrapolation — the pattern may not continue and the model may give unrealistic values.

y = 2 · 4^1.5 = 2 · 8 = 16.

The model is not valid for that input. Negative length, mass, or similar quantities are physically impossible. Either the input is outside the valid domain or the model breaks down.

f(t) = a sin(bt + c) + d. a = amplitude (half the range). Period = 2π/b. c = phase shift. d = midline (vertical shift).

Amplitude = 3. It is the coefficient of sin — the distance from the midline to the maximum or minimum.

Period = 2π ÷ (π/6) = 2π × 6/π = 12.

Midline y = 10. Amplitude = 4, so f oscillates between 10 − 4 = 6 and 10 + 4 = 14.

Amplitude = (max − min) / 2. Midline = (max + min) / 2.

Amplitude = (18 − 4)/2 = 7. Midline = (18 + 4)/2 = 11.

Midline = (8 + 24)/2 = 16°C. Amplitude = (24 − 8)/2 = 8°C.

2π/b = 24 → b = 2π/24 = π/12.

Amplitude = (max − min)/2, not the maximum value alone. If min = 4, amplitude = (18 − 4)/2 = 7, not 18.

Period: how long one complete cycle takes (in time units, e.g. hours). Frequency: cycles per unit time = 1/period.

b is not the period — it is a parameter inside the argument. Period = 2π/b. For b = 2, period = π, not 2.

Maximum = midline + amplitude = 12 + 5 = 17. The midline d is not the maximum.

f(6) = 7 sin(π · 6/12) + 15 = 7 sin(π/2) + 15 = 7(1) + 15 = 22.

h(3) = 3 sin(π/2) + 5 = 3(1) + 5 = 8 m.

Either a calculation error, or the model is being used outside its valid range. A sinusoidal model never exceeds amplitude + midline.

8 sin(πt/12) + 12 = 20 → sin(πt/12) = 1 → πt/12 = π/2 → t = 6 hours.