Curriculum · Course · Week 2 · فارسی
A linear function changes at a constant rate. A quadratic has a turning point. A rational expression can exclude inputs where its denominator vanishes. A sinusoid repeats after one period.
Example 1. For f(x) = (x − 2)² − 1, the vertex is (2, −1). The zeros are 1 and 3 because x − 2 must be ±1.
Example 2. For v(t) = 3 sin(2πt), the amplitude is 3 and the period is 1 s, because the angle advances by 2π each second. At a quarter period the value is 3.
Practice. Find the domain of 1/(x − 3) and the period of 2 sin(4πt). Answers: all real numbers except 3, and 0.5 s.
| Family | What to look for | Engineering example |
|---|---|---|
| Linear | Equal input steps give equal output changes | Ideal sensor calibration |
| Quadratic | A turning point and mirror symmetry about the vertex | Resistor power as a function of current at fixed resistance |
| Rational | A denominator can exclude inputs | A ratio with a forbidden zero denominator |
| Sinusoidal | A repeating pattern with a fixed period | An ideal AC voltage |
For v(t) = 2 + 4 sin(πt), the midline is 2 and the amplitude is 4. Maximum and minimum are 6 and −2. Because πt needs an increase of 2π for a full turn, the period is 2 s. Starting at t = 0, the quarter-cycle samples at 0, 0.5, 1, 1.5, and 2 s are 2, 6, 2, −2, and 2. Sketch them in that order. The vertical shift changes the midline but leaves the period unchanged.
Answers. 1. 3 for the line and ±3 for the quadratic. 2. All real x except 1, with a hole at (1, 2). 3. Midline −1, amplitude 3, period 0.5 s, and v(0) = 2.
The blue curve is sin(2πt) and the orange curve is 2 sin(2πt). Both repeat after 1 second, but the orange curve reaches twice the height above the midline. Multiplying outside the sine changes amplitude. Multiplying the angle inside it changes how quickly the cycle repeats.
Read the graph. Compare the curves at one quarter of the period. The blue curve is at 1 and the orange curve is at 2. Both cross the midline at the same times. If the orange curve were instead 2 sin(4πt), its peak height would stay 2 but it would complete two cycles in the same second.
The purpose of a function family is to constrain a model’s behaviour. A polynomial is defined for every real input, but a quotient or root may not be. A vertical asymptote is not a point on a graph. An excluded input caused by a cancelled factor produces a hole instead. A useful sketch identifies these restrictions before connecting points.
For y=A f(B(x−h))+D, the vertical scale is |A|, the vertical shift is D and the horizontal shift is h. When B is nonzero, the horizontal scale factor is 1/|B|, not |B|. A negative A reflects across the horizontal axis; a negative B reverses the horizontal direction. Determine the transformed input first rather than memorising an arrow diagram.
For r(x)=(x²−4)/(x²−x−2), factor the numerator as (x−2)(x+2) and denominator as (x−2)(x+1). Both x=2 and x=−1 are excluded. Away from them, r(x)=(x+2)/(x+1)=1+1/(x+1). At x=2 the missing value is 4/3, producing a hole. Near x=−1 the remaining denominator approaches zero while its numerator does not, producing a vertical asymptote. The graph approaches the horizontal line y=1 for large input magnitude. Cancellation simplifies a rule without changing its original domain.
Radians measure angle by arc length divided by radius. One revolution is 2π rad=360°, so θrad=θdeg π/180. On the unit circle the point at angle θ is (cos θ,sin θ). Thus sin²θ+cos²θ=1. Also tan θ=sin θ/cos θ where cosine is nonzero. Tangent has period π and vertical asymptotes at π/2+kπ for integer k. Sine and cosine have period 2π.
For a time waveform v(t)=D+A sin(ωt+φ), amplitude is |A|, angular frequency is |ω| rad/s, frequency is f=|ω|/(2π) Hz and period is T=2π/|ω| when ω is nonzero. The phase shift in time is −φ/ω. The angle fed to sine must be dimensionless; writing ω in rad/s makes this interpretation explicit.
Let v(t)=1+2 cos(4πt−π/2) V with t in seconds. The midline is 1 V, amplitude 2 V, frequency 2 Hz and period 0.5 s. Factor the angle as 4π(t−1/8), so the cosine shifts right by 0.125 s. At t=0 the value is 1 V; at t=0.125 s it reaches 3 V. The range is [−1,3] V. Its negative minimum is a polarity relative to the reference, not a negative amplitude.
Problems 1–6 are core; 7–8 are optional extensions. Sketch before checking with software.
Common mistakes: calculator degree mode in a radians problem, confusing frequency with angular frequency, and calling a vertical shift an amplitude. Always label the time axis with seconds when reporting hertz.