Key-point sketches and transformed tangent domains
導入済み| English | 意味 |
|---|---|
| asymptote/ˈæsɪmptəʊt/ | 漸近線 |
A shifted and stretched trig graph can hide its key points. Which input angles locate its peaks, crossings and breaks?
- A shifted and stretched trig graph can hide its key points. Which input angles locate its peaks, crossings and breaks?
- This lesson studies asymptote 渐近线: A line approached by a graph while its values grow without bound or tend toward a limit.
Choose the mathematical structure
- For a sine/cosine graph a f(kx+c)+d, amplitude is |a|, centre line y=d and period 2π/|k| for k≠0. Transform the five quarter-cycle base inputs to locate key points. A negative a reflects vertically. For tangent the period is π/|k|, there is no amplitude and the range is all real numbers; exclude every input giving a base-angle π/2+jπ.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Why does y=tan(2x+π/3) have no amplitude and no joinable branches?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Tangent has no maximum or minimum - "amplitude" is meaningless for it. Each branch rises between asymptotes x=π/12+jπ/2, and joining across an asymptote invents points that do not exist.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For y=2cos(x−π/3)−1, one cycle has points (π/3,1),(5π/6,−1),(4π/3,−3),(11π/6,−1),(7π/3,1). The range is [−3,1] and period 2π. For y=−3sin2x+1, the points x=0,π/4,π/2,3π/4,π give y=1,−2,1,4,1; the period is π and range [−2,4]. For y=tan(2x+π/3), zeros satisfy 2x+π/3=jπ, hence x=jπ/2−π/6. Asymptotes satisfy 2x+π/3=π/2+jπ, hence x=π/12+jπ/2. On 0≤x<π, the zeros are π/3 and 5π/6 and the asymptotes π/12 and 7π/12.
Key-point sketches and transformed tangent domains
For a sine/cosine graph a f(kx+c)+d, amplitude is |a|, centre line y=d and period 2π/|k| for k≠0
Justify each triangle candidate or transformed key point against its defining conditions.
Find the maximum of 2cos(x−π/3)−1.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Cosine can attain 1, giving 2×1−1=1.
Test a tempting shortcut
- The cosine maximum need not occur at x=0 after a horizontal shift. For the stated tangent function, draw separate increasing branches between asymptotes. A negative horizontal or vertical factor reverses the direction; never join branches across a break. Vertical translations change outputs but not the period or excluded x values. A displayed interval may contain only part of a complete shifted cycle.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
A transformed tangent graph has a finite amplitude and can be joined across its vertical asymptotes. This claim is false. Explain which definition or assumption it violates.
Find its minimum.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Cosine can attain −1, giving 2×(−1)−1=−3.
A transformed tangent graph has a finite amplitude and can be joined across its vertical asymptotes.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The cosine maximum need not occur at x=0 after a horizontal shift. For the stated tangent function, draw separate increasing branches between asymptotes. A negative horizontal or vertical factor reverses the direction; never join branches across a break. Vertical translations change outputs but not the period or excluded x values. A displayed interval may contain only part of a complete shifted cycle.
Interpret a new situation
- Find period and centre line first, then map exact base-angle key points. Add enough repeats to cover the requested interval and label scale, extremes and asymptotes. To find x-axis crossings, solve the output equation rather than assuming every centre-line crossing has y=0. Check a sample point in each tangent branch and preserve the original domain.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Write the period of tan(2x+π/3) as kπ. Find k.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Tangent has base period π; dividing by input coefficient 2 gives π/2, hence k=1/2.
Match each transformation feature of a f(kx+c)+d to its effect on a sine graph.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Transform the five quarter-cycle key points; a negative a reflects vertically, and shifts move the maximum away from x=0.
Use this in your course
- 7357 · A-level · E. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
A line approached by a graph while its values grow without bound or tend toward a limit. Choose the relationship, show the method, check its assumptions and interpret the result.