y = a sin k(x + p), y = a cos k(x + p) and y = a tan k(x + p) — amplitude, period, horizontal shift, asymptotes, and reading the graphs to answer questions.
1The three parent graphs
y = cos x · amplitude 1, period 360°
y = tan x · no amplitude, period 180°
2The effect of a — amplitude
Amplitude = |a|. It stretches the graph vertically. A negative a flips it in the x-axis.
3The effect of k — period
k does not change the amplitude and a does not change the period. Keep the two effects separate.
4The effect of p — horizontal shift
y = cos(x + 60°) moves left by 60°; y = cos(x − 60°) moves right. The sign inside the bracket is the opposite of the direction you might expect.
5The tangent graph
tan has asymptotes every 180°, starting at 90°, and no maximum or minimum — so it has no amplitude and its range is y ∈ ℝ.
6Reading a trig graph
- 1Amplitude = |2| = 2.The graph reaches 2 and −2.
- 2Period = 360°/1 = 360°.k = 1.
- 3Range: −2 ≤ y ≤ 2.Between the two extremes, endpoints included.
- 1Period = 360°/2 = 180°.
- 2360° ÷ 180° = 2 complete cycles.A full cycle every 180° fits exactly twice into 360°.
Always write the period and amplitude down before you draw. Then plot the key points every quarter-period — that is where the maxima, minima and x-intercepts sit.
7Two parameters at once: y = a sin k(x + p)
The ATP asks for sketches with at most two parameters at a time. Take them in a fixed order and the graph builds itself.
amplitude = |a| · period = 360°/(|k|) (tan: 180°/(|k|)), k ≠ 0 · horizontal shift = p units LEFT
- a stretches it vertically. Negative a flips it in the x-axis.
- k squashes it horizontally. k = 2 halves the period to 180°, so you get two full waves in 360°.
- p slides it sideways — left for positive p, right for negative p, once k has been taken out as a factor.
Mind the bracket. y = a sin(kx + p) is NOT shifted p units — take the k out first: a sin k(x + p/k), so the shift is p/k. When k = 1 the two forms are the same, which is why most Grade 11 questions are written as y = a sin(x + p).
- 1Amplitude = 2, so the curve runs from −2 to 2.|a| = 2
- 2k = 1, so the period is still 360°.One complete wave across the interval.
- 3p = −30°, so the whole graph moves 30° to the RIGHT.Negative p shifts right.
- 4The zeros move from 0°, 180°, 360° to 30°, 210° and 390° — so inside [0° ; 360°] they are 30° and 210° (390° lies outside the interval).Add 30° to every key x-value, then keep only the ones in the interval.
- 5The maximum moves from 90° to 120°, and the minimum from 270° to 300°.Same shift applies to the turning points.
- 1Amplitude = 3
- 2Period = 360° ÷ 2 = 180°k = 2 halves the period.
- 3A cosine graph is at a maximum at the start of each period.
- 4Maxima at x = 0°, 180° and 360°.Two full waves fit in the interval.
Multiplying the period by k instead of dividing. k = 2 makes the wave faster, so the period gets smaller: 360° ÷ 2 = 180°. If your period came out bigger than 360° for k > 1, you have divided the wrong way round.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Write down the amplitude of y = 2 sin x.
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2 - 2Write down the period of y = sin 2x.
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360°/2 = 180° - 3Write down the range of y = 2 sin x.
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−2 ≤ y ≤ 2 - 4Describe the transformation from y = cos x to y = cos(x + 60°).
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A shift of 60° to the left - 5Write down the period of y = tan 2x.
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180°/2 = 90° - 6Write down the equations of the asymptotes of y = tan x for x ∈ [0° ; 360°].
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x = 90° and x = 270° - 7How many complete cycles does y = sin 2x complete in [0° ; 360°]?
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2 - 8Write down the range of y = tan x.
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y ∈ ℝ - 9For y = 4 sin 3x, write down the amplitude and the period, and state how many complete waves appear in [0° ; 360°].
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Amplitude 4 ✓; period = 360° ÷ 3 = 120° ✓; three complete waves ✓ - 10Describe fully the transformation from y = cos x to y = −2 cos(x + 45°).
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Amplitude becomes 2 ✓; the graph is reflected in the x-axis because a is negative ✓; and it shifts 45° to the LEFT ✓ - 11The graph of y = a sin kx, where a > 0 and k > 0, has amplitude 5 and period 90°. Determine a and k, and write down the equation.
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a = 5 ✓; 360° ÷ k = 90° so k = 4 ✓; y = 5 sin 4x ✓
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