LEARN · EXPLAIN · REVISE

Describe the tangent and normal to y=x² at zero.

Read the idea, work independently, then explain what changed.

Return to the lesson / paper ↗

高二選擇性必修第二册(A版).pdf · 5.1 · PDF 64 / printed page 59

TOPIC 01

Rates and tangent meaning

Derive tangent slope from shrinking secants, then connect slope, point and local approximation.

What you will be able to explain

Derive tangent slope from shrinking secants, then connect slope, point and local approximation.
Distinguish average rate, instantaneous rate, height, normal and inverse rate.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Standard#Your turn

Describe the tangent and normal to y=x² at zero.

y=x^2;\ a=0

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Treat a vertical normal with its equation, not a finite slope.

Hint 2

Start from this intermediate relation.

f'(0)=0

Worked solution

Treat a vertical normal with its equation, not a finite slope.
Establish and simplify this mathematical relation.
$f'(0)=0$
Establish and simplify this mathematical relation.
$T:y=0;\ N:x=0$
The expression −1/0 is undefined; the normal is still a valid vertical line.

Tangent y=0; normal x=0.

Checks and common pitfalls: The expression −1/0 is undefined; the normal is still a valid vertical line.

Reasoning checklist · self / teacher assessment

Treat a vertical normal with its equation, not a finite slope.
Give the intermediate relations and explain the conclusion with its restrictions.
The expression −1/0 is undefined; the normal is still a valid vertical line.

Think first. Reveal a hint when the class is ready.

Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Predict	Predict slopes of secants to x² at a=1 from both sides.
8	Build the idea	Sample response: quotient 2+h tends to 2; h cannot initially be zero.
10	Guided comparison	Sample response: slope 4 at (2,4) gives y−4=4(x−2); normal slope is −1/4.
12	Independent practice and discussion	Contrast the corner task with normal and local-error tasks; require a reason before calculation.
5	Exit and next step	Exit: explain why f(3)=9 but f′(3)=6, and when a normal is vertical.

Question sequence

Predict slopes of secants to x² at a=1 from both sides.
Sample response: slope 4 at (2,4) gives y−4=4(x−2); normal slope is −1/4.

Board development

From secant to tangent
$f\prime(a)=\lim_{h\to0}[f(a+h)-f(a)]/h$

Expected correct and incorrect responses

Correct sample: 6
Anticipated wrong response: 9 — Substitutes the function height for its derivative.
Anticipated wrong response: Derivative is 0. — Applies a differentiable interior-minimum theorem to a corner.
Confusing f(a) with f′(a).
Treating a corner as a horizontal tangent.

Follow-up questions

Follow-up: Differentiate the function, then evaluate its derivative. Then explain why 9 fails.
Follow-up: Compare left and right difference quotients. Then explain why Derivative is 0. fails.
Relate the vertical gap h² to the accuracy of the local approximation.

Remediation

Sample response: quotient 2+h tends to 2; h cannot initially be zero.
Return to the first example, identify the missing condition, then solve one fresh task without viewing its solution.

Exit check

Exit: explain why f(3)=9 but f′(3)=6, and when a normal is vertical.

Assessment checklist and sample responses

Representation and conditions
For h≠0, [(1+h)²−1]/h=2+h; its limit is 2.
Connected mathematical reasoning
At (2,4), f′(2)=4, so y−4=4(x−2).
Interpretation and boundary check
The exact minus predicted value is h²≥0, even for negative h.