Practise questions · Salam Academy

SA-9709-P1-FN5-0011 markL1 FoundationCalculatorDescribe

The transformation that maps

y=f(x)

onto

y=f(x+4)

is

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0021 markL1 FoundationCalculatorDescribe

The transformation that maps

y=f(x)

onto

y=3f(x)

is

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-G012 marksL1 FoundationCalculatorImage of a point

f(x)=x^2

. State the coordinates of the vertex of

y=f(x-3)+2

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-Q011 markL1 FoundationCalculatorDescribe

The graph of

y=f(x)-5

is the graph of

y=f(x)

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-Q021 markL1 FoundationCalculatorDescribe

y=f(-x)

is a reflection of

y=f(x)

in

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-Q031 markL1 FoundationCalculatorDescribe

A stretch parallel to the

x

-axis with scale factor

\tfrac13

maps

y=f(x)

onto

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0031 markL2 StandardCalculatorDescribe

The transformation that maps

y=f(x)

onto

y=f(2x)

is a stretch parallel to the

x

-axis with scale factor

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0042 marksL2 StandardCalculatorImage of a point

The point

(6,4)

lies on

y=f(x)

. Find its image on

y=f(2x)

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0052 marksL2 StandardCalculatorImage of a point

The point

(6,4)

lies on

y=f(x)

. Find its image on

y=3f(x)-1

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0073 marksL2 StandardCalculatorNew equation

f(x)=x^2-4x

. The graph of

y=f(x)

is translated by

\dbinom{1}{3}

. Find the equation of the new graph in the form

y=x^2+bx+c

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0082 marksL2 StandardCalculatorImage of a point

The graph of

y=x^2

is transformed to

y=2(x+3)^2-5

. State the coordinates of the vertex of the new graph.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0091 markL2 StandardCalculatorCombinations

The graph of

y=f(x)

is reflected in the

x

-axis and then translated by

\dbinom{0}{2}

. The equation of the new graph is

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0102 marksL2 StandardCalculatorImage of a point

The point

(1,-3)

lies on

y=f(x)

. Find its image on

y=f(x-2)+5

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0122 marksL2 StandardCalculatorDescribe

The graph of

y=\sin x

is transformed to

y=3\sin2x

, with

x

in degrees. State (i) the maximum value of

3\sin2x

(ii) its period in degrees.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0142 marksL2 StandardCalculatorImage of a point

f(x)=x^3

. The point

(2,8)

lies on

y=f(x)

. Find its image on

y=-f(x-1)

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0062 marksL3 ExamCalculatorImage of a point

The point

(6,4)

lies on

y=f(x)

. Find its image on

y=-f(x+2)

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0111 markL3 ExamCalculatorCombinations

y=f(2x+6)

is obtained from

y=f(x)

by a stretch parallel to the

x

-axis with scale factor

\tfrac12

, followed by a translation by

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-FN5-0135 marksL5 ExtensionCalculatorNew equation

The curve

y=x^2-6x+10

is the image of

y=x^2

under a translation by

\dbinom{p}{q}

. (i) Find

p

and

q

. (ii) The curve

y=x^2-6x+10

is then stretched parallel to the

y

-axis with scale factor

2

. Find the equation of the final curve.

Salam Academy Original Practice · Proposed mark scheme

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