Sketching Cubic Graphs (Grade 12) — Worksheet & Memo

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Grade 12 · Calculus

Sketching Cubic Graphs (Grade 12)

MARKING GUIDELINE

Marks

Duration

50 minutes

Questions

Name:

Class:

Date:

Mark

/ 32

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
At a STATIONARY point of a cubic …
(1)
A)f'(x) = 0✓
B)f(x) = 0
C)f''(x) = 0
D)x = 0
Answer: A — A stationary point is where the tangent is horizontal.
- B — that is an x-INTERCEPT
- C — that is a point of INFLECTION
- D — x = 0 is on the y-axis, which has nothing to do with it
1.2
h(x) = ax³ + bx² has turning points at (0 ; 0) and (4 ; 32). Then …
(1)
A)a = −1 and b = 6✓
B)a = 1 and b = −6
C)a = −1 and b = −6
D)a = 6 and b = −1
Answer: A — h'(x) = 3ax² + 2bx is zero at x = 4, and h(4) = 32; solving both gives a = −1 and b = 6.
- B — every sign has been reversed, which would put the turning point below the axis
- C — b must be positive for (4 ; 32) to be a maximum
- D — the values of a and b have been swapped
1.3
For h(x) = −x³ + 6x², the x-intercept other than the origin is at …
(1)
A)x = 6✓
B)x = 4
C)x = −6
D)x = 32
Answer: A — −x³ + 6x² = x²(6 − x), which is zero at x = 0 and at x = 6.
- B — x = 4 is a TURNING point, not an x-intercept
- C — 6 − x = 0 gives x = +6
- D — 32 is the y-value at the turning point, not an x-intercept
1.4
For h(x) = −x³ + 6x², h is INCREASING for …
(1)
A)0 < x < 4✓
B)x < 0 or x > 4
C)x > 4
D)x < 0
Answer: A — h'(x) = 3x(4 − x), which is positive between its roots 0 and 4.
- B — that is where h is DECREASING
- C — beyond x = 4 the graph falls
- D — below x = 0 the graph falls
1.5
For h(x) = −x³ + 6x², h is CONCAVE DOWN for …
(1)
A)x > 2✓
B)x < 2
C)x > 4
D)0 < x < 4
Answer: A — h''(x) = −6x + 12, which is negative when x > 2.
- B — that is where h is concave UP
- C — 4 is a turning point, not the point of inflection
- D — that is where h is INCREASING, which is a different question
1.6
The POINT OF INFLECTION of a cubic is where …
(1)
A)f''(x) = 0✓
B)f'(x) = 0
C)f(x) = 0
D)the graph cuts the y-axis
Answer: A — The concavity changes there, so the second derivative is zero.
- B — that is a STATIONARY point
- C — that is an x-intercept
- D — the y-intercept has nothing to do with concavity
1.7
A cubic whose coefficient of x³ is POSITIVE …
(1)
A)rises to the right and falls to the left✓
B)falls to the right and rises to the left
C)is increasing everywhere
D)has no turning points
Answer: A — With a > 0 the graph comes up from below on the left and heads upward on the right.
- B — that describes a cubic with a NEGATIVE coefficient
- C — a cubic with two turning points decreases between them
- D — a cubic may have two turning points, or none
1.8
How many turning points can a cubic graph have?
(1)
A)at most two✓
B)exactly two
C)exactly one
D)at most three
Answer: A — f'(x) is a quadratic, so it has at most two roots.
- B — y = x³ has none at all
- C — exactly one would need f' to have a single root, which is the inflection case
- D — f' is a quadratic and cannot have three roots
1.9
For f(x) = x³ − 3x², the LOCAL MINIMUM is at …
(1)
A)(2 ; −4)✓
B)(0 ; 0)
C)(2 ; 4)
D)(−2 ; −20)
Answer: A — f'(x) = 3x(x − 2) is zero at x = 0 and x = 2, and f''(2) = 6 > 0 makes x = 2 the minimum.
- B — that is the local MAXIMUM, where f''(0) = −6 < 0
- C — a sign slip: 8 − 12 = −4, not +4
- D — x = −2 is not a stationary point at all
1.10
To decide whether a stationary point is a maximum or a minimum, you can …
(1)
A)test the sign of f''(x) at that point✓
B)test the sign of f(x) at that point
C)check whether x is positive
D)count the x-intercepts
Answer: A — A positive second derivative means concave up, and hence a minimum.
- B — the VALUE of f says nothing about the shape there
- C — the sign of x is irrelevant
- D — the number of intercepts does not identify a turning point's nature

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[12 MARKS]

Given f(x) = −x³ + 6x².

2.1
Calculate the coordinates of the x-intercepts of f.
(4)
x²(6 − x) = 0 (2)
x = 0 (a double root) or x = 6 (1)
(0 ; 0) and (6 ; 0) (1)
2.2
Determine the coordinates of the stationary points of f.
(5)
f′(x) = −3x² + 12x (1)
−3x(x − 4) = 0 (2)
x = 0 gives (0 ; 0) (1); x = 4 gives (4 ; 32) (1)
2.3
Determine the coordinates of the point of inflection.
(3)
f″(x) = −6x + 12 = 0 (1)
x = 2 (1)
f(2) = 16, so (2 ; 16) (1)

Question 3

[10 MARKS]

Answer the questions below about the same graph.

3.1
Write down the values of x for which f is concave up.
(2)
f″(x) = −6x + 12 > 0 (1)
x < 2 (1)
3.2
Determine the equation of the tangent to f at x = 1.
(4)
f′(1) = −3 + 12 = 9 (1)
f(1) = 5, so the point is (1 ; 5) (1)
5 = 9(1) + c gives c = −4 (1)
y = 9x − 4 (1)
3.3
For which values of k will −x³ + 6x² = k have THREE distinct real roots? Justify your answer.
(4)
The horizontal line y = k must cut the graph three times (1)
That happens strictly between the two turning-point values (1)
0 < k < 32 (2)

TOTAL: 32 marks

This question paper consists of 3 questions.