Functions & Graphs

7 methods · 29 questions

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Translations, reflections and stretches act on the equation by substitution: a shift by $a$ in $x$ replaces $x$ by $x - a$ , a reflection in the $y$ -axis replaces $x$ by $- x$ , a reflection in the $x$ -axis replaces $y$ by $- y$ , a horizontal stretch by factor $s$ replaces $x$ by $x / s$ , a vertical shift adds a constant. Apply the moves in the stated order to get the new equation, then read its vertex or features off. Crucially, a horizontal shift or a reflection neither creates nor destroys $x$ -intercepts (and a pure translation cannot change the leading coefficient), whereas a vertical shift or stretch can change the number of roots, so transformation claims can be accepted or rejected almost on sight.

Trigger: the question describes shifts, reflections or stretches of $y = f (x)$ , or asks whether a given curve is such a transform.

Instances:
(i) compose the substitutions in order and expand to obtain the resulting equation;
(ii) track only the image of the vertex or stationary point rather than the whole curve;
(iii) declare a horizontal shift or reflection root-count-preserving, and match coefficients knowing a translation leaves the leading coefficient fixed.

Linked questions (3)

TMUA 2016 Paper 2 · Q11

f (x)

is a polynomial with real coefficients.

The equation

f (x) = 0

has exactly two real roots,

x = - p

and

x = p

, where

p > 0

.

Consider the following three statements:
1.

f^{'} (x) = 0

for exactly one value of

x

between

- p

and

p

2. The area between the curve

y = f (x)

, the

x

-axis and the lines

x = - p

and

x = p

is given by

2 \int_{0}^{p} f (x) d x

3. The graph of

y = - f (- x)

intersects the

x

-axis at the points

x = - p

and

x = p

only

Which of the above statements must be true?

Anone
B1 only
C2 only
D3 only
E1 and 2 only
F1 and 3 only
G2 and 3 only
H1, 2 and 3

Worked solution

Answer:D

Statement 1: A polynomial with roots only at

\pm p

could have multiple stationary points between

- p

and

p

without crossing the axis (e.g. a degree-6 polynomial dipping and rising multiple times). False in general. Statement 2: The area formula

2 \int_{0}^{p} f (x) d x

requires

f

to be symmetric about

x = 0

(even function) and non-negative on

[0, p]

. Neither is guaranteed. For example

f (x) = (x + 1) (x - 1) (x^{2} + x + 1)

has roots

\pm 1

but is not symmetric, so

\int_{- 1}^{1} f (x) d x \neq = 2 \int_{0}^{1} f (x) d x

. False in general. Statement 3: The graph of

y = - f (- x)

is obtained by reflecting

y = f (x)

in both the

y

-axis and the

x

-axis (i.e. rotation by

18 0^{\circ}

about the origin). Reflecting in the

y

-axis maps roots

\pm p

\mp p

, still

\pm p

. Reflecting in the

x

-axis does not change which

x

-values are roots. So

y = - f (- x)

intersects the

x

-axis exactly at

x = \pm p

. Must be true. Hence the answer is D.

TMUA 2018 Paper 2 · Q15

It is given that

f (x) = x^{3} + 3 q x^{2} + 2

, where

q

is a real constant.

The equation

f (x) = 0

has 3 distinct real roots.

Which of the following statements is/are necessarily true?

I. The equation

f (x) + 1 = 0

has 3 distinct real roots.

II. The equation

f (x + 1) = 0

has 3 distinct real roots.

III. The equation

f (- x) - 1 = 0

has 3 distinct real roots.

Anone of them
BI only
CII only
DIII only
EI and II only
FI and III only
GII and III only
HI, II and III

Worked solution

Answer:G

The stationary points of

f (x) = x^{3} + 3 q x^{2} + 2

satisfy

f^{'} (x) = 3 x^{2} + 6 q x = 0

, giving

x = 0

x = - 2 q

. The stationary values are

f (0) = 2

and

f (- 2 q) = 2 + 4 q^{3}

.

For

f (x) = 0

to have 3 distinct roots, the stationary values must have opposite signs. Since

f (0) = 2 > 0

, we need

2 + 4 q^{3} < 0

, i.e.

q^{3} < - \frac{1}{2}

.

Statement I:

y = f (x) + 1

shifts the graph up by 1. The new stationary values are

3

and

3 + 4 q^{3}

. Since

2 + 4 q^{3} < 0

, we have

3 + 4 q^{3} < 1

, which could be positive or negative. If

- 1 < 4 q^{3} + 2 < 0

, then

3 + 4 q^{3}

is between 2 and 3, which is positive, meaning

y = f (x) + 1

has only one root. So I is not necessarily true.

Statement II:

y = f (x + 1)

is a horizontal shift, which does not change the number of real roots. So II is necessarily true.

Statement III:

y = f (- x) - 1

is a reflection in the

y

-axis followed by a downward shift of 1. Reflection does not change root count. The new stationary values are

2 - 1 = 1 > 0

(at the local max after reflection) and

(2 + 4 q^{3}) - 1 = 1 + 4 q^{3}

. Since

q^{3} < - \frac{1}{2}

, we have

4 q^{3} < - 2

, so

1 + 4 q^{3} < - 1 < 0

. The two stationary values have opposite signs, guaranteeing 3 distinct roots. So III is necessarily true.

Statements II and III are necessarily true. Hence the answer is G.

TMUA 2019 Paper 2 · Q20

When the graph of the function

y = f (x)

, defined on the real numbers, is reflected in the

y

-axis and then translated by 2 units in the negative

x

-direction, the result is the graph of the function

y = g (x)

.

When the graph of the same function

y = f (x)

is translated by 2 units in the negative

x

-direction and then reflected in the

y

-axis, the result is the graph of the function

y = h (x)

.

Which one of the following conditions on

y = f (x)

is necessary and sufficient for the functions

g (x)

and

h (x)

to be identical?

A $f (x) = f (x + 2)$ for all $x$
B $f (x) = f (x + 4)$ for all $x$
C $f (x) = f (x + 8)$ for all $x$
D $f (x) = f (- x)$ for all $x$
E $f (x) = f (2 - x)$ for all $x$
F $f (x) = f (4 - x)$ for all $x$
G $f (x) = f (8 - x)$ for all $x$

Worked solution

Answer:B

We derive

g

and

h

algebraically.

For

g (x)

: Reflect

f (x)

in the

y

-axis to get

f (- x)

. Then translate by 2 in the negative

x

-direction (replace

x

x + 2

g (x) = f (- (x + 2)) = f (- x - 2)

.

For

h (x)

: Translate

f (x)

by 2 in the negative

x

-direction to get

f (x + 2)

. Then reflect in the

y

-axis (replace

x

- x

h (x) = f (- x + 2)

.

Condition for

g = h

: We need

f (- x - 2) = f (- x + 2)

for all

x

. Substituting

u = - x - 2

(so

- x + 2 = u + 4

), this becomes

f (u) = f (u + 4)

for all

u

, i.e.

f (x) = f (x + 4)

for all

x

.

The correct answer is B.