A linear model $y = mx + c$ describes a constant rate of change. In context, the gradient $m$ is the rate of change of $y$ with respect to $x$, and the $y$-intercept $c$ is the initial value.
$f(x) = mx + c$
$m = \dfrac{\Delta y}{\Delta x}$ = gradient = rate of change
$c$ = $y$-intercept = value when $x=0$
Linear models are appropriate when a scatter diagram shows a roughly linear trend and the rate of change is constant throughout the domain. Always assess whether a linear model is suitable before fitting one.
Quadratic models $f(x) = ax^2 + bx + c$ appear in projectile motion, profit/cost analysis, and engineering. The vertex gives the maximum or minimum value.
Vertex form: $f(x) = a(x-h)^2 + k$, vertex at $(h,k)$
Vertex coordinates: $h = -\dfrac{b}{2a}$, $k = f(h)$
$a > 0$: opens upward (minimum at vertex)
$a < 0$: opens downward (maximum at vertex)
A company's weekly profit is modelled by $P(x) = -2x^2 + 80x - 300$ (dollars), where $x$ is the number of units produced (in hundreds). Find the maximum profit and the production level that achieves it.
Exponential functions model growth and decay processes where the rate of change is proportional to the current quantity.
General form: $y = a \cdot b^x$ where $a$ = initial value, $b$ = growth/decay factor
$b > 1$: exponential growth $0 < b < 1$: exponential decay
Continuous form: $y = ae^{kx}$ where $k > 0$ (growth) or $k < 0$ (decay)
Half-life: Time for quantity to halve: $t_{1/2} = \dfrac{\ln 2}{k}$ (for $y=ae^{-kt}$)
A drug concentration in the bloodstream is modelled by $C(t) = 80e^{-0.2t}$ mg/L, where $t$ is in hours. Find: (a) the initial concentration; (b) the concentration after 3 hours; (c) when concentration falls below 10 mg/L.
Logistic growth occurs when a population is limited by a carrying capacity $L$. Unlike exponential growth, logistic growth slows as the population approaches $L$.
$$P(t) = \frac{L}{1+Ce^{-kt}}$$
$L$ = carrying capacity (maximum sustainable population)
As $t \to \infty$, $P \to L$ (the population approaches but never exceeds $L$).
Growth is fastest when $P = L/2$ (the inflection point).
A fish population follows $P(t) = \dfrac{5000}{1+9e^{-0.3t}}$ where $t$ is in years. Find the carrying capacity, initial population, and when the population exceeds 4000.
(a) $T(0) = 20 + 65 = 85$°C
(b) As $t\to\infty$: $T\to 20$°C (room temperature)
(c) $20+65e^{-0.1t}=40 \Rightarrow e^{-0.1t}=20/65=4/13 \Rightarrow t = -10\ln(4/13) = 10\ln(13/4) \approx 11.9$ min
Vertex: $t = -\dfrac{20}{2(-5)} = 2$ s. Max height $= -5(4)+40+2 = 22$ m.
Hits ground when $h=0$: $5t^2-20t-2=0 \Rightarrow t = \dfrac{20\pm\sqrt{400+40}}{10} \approx \dfrac{20+\sqrt{440}}{10} \approx 4.10$ s
Each $y$ value is multiplied by 3 as $x$ increases by 1 (constant ratio). This is exponential growth.
$y = a \cdot b^x$. At $x=0$: $a=2$. Growth factor $b=3$. Equation: $y = 2 \times 3^x$.