Sketching the Parabola: The Complete Exam Guide — Worksheet & Memo

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Grade 10 · Functions

Sketching the Parabola: The Complete Exam Guide

MARKING GUIDELINE

Marks

Duration

55 minutes

Questions

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Class:

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Mark

/ 35

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
The axis of symmetry of y = x² − 4x + 3 is …
(1)
A)x = 2✓
B)x = −2
C)x = 4
D)x = 3
Answer: A — x = −b ÷ (2a) = 4 ÷ 2 = 2.
- B — dropped the minus in −b, so the sign came out wrong
- C — used −b on its own — the 2a was never divided out
- D — read off the constant term instead
1.2
The turning point of y = x² − 4x + 3 is …
(1)
A)(2 ; −1)✓
B)(2 ; 3)
C)(−2 ; 15)
D)(2 ; 1)
Answer: A — At x = 2, y = 4 − 8 + 3 = −1.
- B — substituted x = 0 instead of x = 2, which gives the Y-INTERCEPT
- C — used x = −2 instead of x = 2
- D — a sign slip: 4 − 8 + 3 is −1, not +1
1.3
The x-intercepts of y = x² − 4x + 3 are …
(1)
A)x = 1 and x = 3✓
B)x = −1 and x = −3
C)x = 3 only
D)x = 4 and x = 3
Answer: A — x² − 4x + 3 = (x − 1)(x − 3), so y = 0 at x = 1 and x = 3.
- B — the factors are (x − 1)(x − 3), so the roots are POSITIVE
- C — 3 is the y-intercept value as well, but there are TWO x-intercepts
- D — read off b and c instead of solving
1.4
The graph of y = −2x² + 5 …
(1)
A)has a maximum turning point✓
B)has a minimum turning point
C)is a straight line
D)has no turning point
Answer: A — a = −2 is negative, so the parabola opens downward and its turning point is a maximum.
- B — a minimum needs a POSITIVE value of a
- C — an x² term makes it a parabola, not a line
- D — every parabola has exactly one turning point
1.5
The range of y = x² + 2 is …
(1)
A)y ≥ 2✓
B)y ≤ 2
C)y ≥ 0
D)y ∈ R
Answer: A — x² is never negative, so the smallest value y can take is 2.
- B — the graph opens UPWARD, so 2 is a minimum, not a maximum
- C — that is the range of y = x², before the shift
- D — a parabola never takes every real y-value
1.6
For y = a(x − p)² + q, the turning point is …
(1)
A)(p ; q)✓
B)(−p ; q)
C)(p ; −q)
D)(q ; p)
Answer: A — The bracket is zero when x = p, and then y = q.
- B — the bracket reads (x − p), so the turning point is at +p
- C — q is ADDED, so it keeps its sign
- D — the coordinates are the wrong way round
1.7
If the discriminant of a quadratic is NEGATIVE, its parabola …
(1)
A)does not cut the x-axis✓
B)touches the x-axis once
C)cuts the x-axis twice
D)is a straight line
Answer: A — A negative discriminant means no real roots, so the curve never reaches y = 0.
- B — that happens when the discriminant is exactly ZERO
- C — that happens when the discriminant is POSITIVE
- D — a discriminant belongs to a quadratic, whose graph is always a parabola
1.8
y = x² − 6x + 8 written in turning-point form is …
(1)
A)y = (x − 3)² − 1✓
B)y = (x − 3)² + 8
C)y = (x − 6)² − 28
D)y = (x + 3)² − 1
Answer: A — Half of −6 is −3, and (x − 3)² = x² − 6x + 9, so 8 − 9 = −1 must be written back.
- B — forgot to subtract the 9 that completing the square puts in
- C — the bracket takes HALF of −6, which is −3, not −6
- D — the sign inside the bracket follows b's sign, so it is −3
1.9
A parabola cuts the x-axis at 2 and at 6 and passes through (0 ; 12). Its equation is …
(1)
A)y = (x − 2)(x − 6)✓
B)y = (x + 2)(x + 6)
C)y = 12(x − 2)(x − 6)
D)y = (x − 2)(x − 6) + 12
Answer: A — y = a(x − 2)(x − 6), and (0 ; 12) gives 12a = 12, so a = 1.
- B — the roots are +2 and +6, so the factors are (x − 2) and (x − 6)
- C — took a = 12 without dividing by (−2)(−6) = 12
- D — added the y-intercept on instead of using it to find a
1.10
A parabola has x-intercepts −1 and 5. Its axis of symmetry is …
(1)
A)x = 2✓
B)x = 4
C)x = 3
D)x = −3
Answer: A — The axis of symmetry lies halfway between the roots: (−1 + 5) ÷ 2 = 2.
- B — used 5 − (−1) = 6 or 5 − 1 = 4 without halving properly
- C — used (1 + 5) ÷ 2 and lost the minus on the −1
- D — used (−1 − 5) ÷ 2 instead of adding the roots

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

[13 MARKS]

Given f(x) = x² − 4x + 3.

2.1
Write down the shape of the graph and give a reason.
(2)
It opens UPWARDS (1) — a = 1, which is positive (reason 1)
2.2
Calculate the coordinates of the x-intercepts and the y-intercept.
(5)
y = 0: (x − 1)(x − 3) = 0 (2)
x-intercepts (1 ; 0) and (3 ; 0) (2)
x = 0: y = 3, so (0 ; 3) (1)
2.3
Calculate the coordinates of the turning point.
(4)
x = −b2a = 42 = 2 (2)
y = 4 − 8 + 3 = −1 (1)
(2 ; −1) (1)
2.4
Write down the range of f.
(2)
y ≥ −1 (2)

Question 3

[12 MARKS]

Answer the questions below about parabolas.

3.1
Determine, using the discriminant, whether y = x² + x + 4 cuts the x-axis.
(4)
Δ = b² − 4ac = 1 − 16 (2) = −15 (1)
Δ < 0, so it does NOT cut the x-axis (1)
3.2
For y = −2x² + 8x − 6, determine the coordinates of the turning point.
(4)
x = −82(−2) = 2 (2)
y = −8 + 16 − 6 = 2 (1)
(2 ; 2) (1)
3.3
Explain what happens to the graph of y = x² when q is changed in y = x² + q, and what happens when a is made negative.
(4)
q shifts the whole parabola vertically (2)
A negative a flips it so that it opens downwards (2)

TOTAL: 35 marks

This question paper consists of 3 questions.