Transformations of Graphs: past paper questions & worked solutions

The minimum point of \(y=f(x)\) is \[ (8,12) \]

For \[ y=f(x)+5 \] the graph moves up by 5: \[ (8,12)\to(8,17) \]

For \[ y=f(2x) \] the graph is horizontally compressed by scale factor \(\frac12\), so the \(x\)-coordinate is halved: \[ (8,12)\to(4,12) \]

The minimum point of \(y=f(x)\) is \((6,-3)\).

For

the graph moves 10 units up:

For

the \(x\)-coordinate is divided by 3:

The minimum point of \(y=f(x)\) is \((5,-4)\).

For

the graph moves 7 units left:

For

the graph moves 6 units down:

The turning point of \(y=f(x)\) is \((6,5)\).

For

the graph moves 4 units right:

For

the \(x\)-coordinate is divided by 3:

The maximum point of \(y=f(x)\) is \((s,t)\).

For

the graph moves 2 units right:

For

the \(y\)-coordinate is multiplied by 3:

The maximum point of \(y=f(x)\) is \((5,7)\).

For

the graph moves 9 units left:

For

the graph moves 3 units up:

The maximum point of \(y=f(x)\) is \((-3,4)\).

For

the graph moves 6 units down:

For

the \(x\)-coordinate is divided by 2:

The turning point of \(y=f(x)\) is \((4,3)\).

For

the graph moves 5 units left:

For

the graph moves 7 units up:

For

the \(x\)-coordinate is divided by 2:

The turning point of \(y=f(x)\) is \((-6,9)\).

For

the \(x\)-coordinate is divided by 3:

For part (b), a point \((4,-5)\) on \(y=g(x)\) maps to \((1,-16)\) on

For the \(x\)-coordinate:

For the \(y\)-coordinate:

The minimum point of \(y=f(x)\) is \[ (5,4) \]

(i) For \[ y=f(x+5) \]

the graph moves 5 units left, so the minimum point becomes \[ (0,4) \]

(ii) For \[ y=3f(x) \]

the y-coordinate is multiplied by 3, so the minimum point becomes \[ (5,12) \]

(iii) For \[ y=f(x)-7 \]

the graph moves 7 units down, so the minimum point becomes \[ (5,-3) \]

For part (a),

The \(x\)-coordinates stay the same and the \(y\)-coordinates change sign.

So the transformation is a reflection in the \(x\)-axis.

For part (b),

First, \(x-1\) moves the graph 1 unit right. Then the factor 2 stretches the graph vertically by scale factor 2.

Use points from the original graph:

Draw the transformed parabola through these points.

The graph has maximum 3 and minimum -3, so \(|a|=3\). Since \(a<0\), \(a=-3\).

The amplitude is half the vertical range.

At \(x=0\), \(y=-3\), so \(-3\sin b^\circ=-3\) and \(\sin b^\circ=1\). Hence \(b=90\) for \(0<b<360\).

The minimum at x=0 fixes the phase when the amplitude is negative.

To use the original input \(p\), set \(2x=p\), giving \(x=p/2\).

The replacement of x by 2x halves horizontal coordinates.

The corresponding output is \(f(p)+5=q+5\).

Adding 5 translates every output upwards by 5.

(a)

For \(y=f(x-4)\), move the graph \(4\) units right:

For \(y=f(3x)\), divide the \(x\)-coordinate by \(3\):

(b) For \(y=g(2x)-3\), halve each \(x\)-coordinate and subtract \(3\) from each \(y\)-coordinate.

Join these points with straight line segments.

For part (a),

Since

So the graph has moved 4 units left and 6 units up.

For part (b), the graph has maximum value \(3\) and minimum value \(-3\), so

From the graph, the minimum is at \(x=135\), so

Point \(A=(-3,2)\) lies on \(y=f(x)\).

For

the graph moves 3 units down:

For

the \(x\)-coordinate is multiplied by 2:

For part (b), point \(B=(p,q)\) lies on \(y=g(x)\).

For

\(x-c\) moves the graph \(c\) units right, and the minus sign reflects the graph in the \(x\)-axis.

So

The minimum point of \(y=f(x)\) is \((-9,15)\).

For

the graph moves 3 units left:

For

the \(y\)-coordinate is divided by 3:

For part (b), the cosine graph has maximum \(2\) and minimum \(-2\), so

The graph has a minimum at \(x=90^\circ\). For \(a\cos(x+b)^\circ\), a minimum occurs when

So

The maximum point of \(y=f(x)\) is \((4,6)\).

For

the graph moves 4 units left:

For

the \(x\)-coordinate is divided by 2:

For part (b),

So the graph moves 4 units right and 6 units up.

Replace \(x\) by \(x-4\), then add 6:

For part (a), the graph has maximum \(2.5\) and minimum \(-2.5\), so

The maximum occurs at \(x=60^\circ\). For \(y=a\cos(x+b)^\circ\), a maximum occurs when

So

For part (b), the minimum point of \(y=f(x)\) is \((4,5)\).

For

the \(x\)-coordinate is divided by 2:

For

the \(y\)-coordinate decreases by 7:

For \(y=3f(x)\), x-coordinates are unchanged and y-coordinates are multiplied by \(3\). Thus \((-5,4)\mapsto(-5,12)\).

The vertical scale factor is positive, so a minimum remains a minimum.

For \(Q(x)=2-(x+3)^2\), \[Q(-x)=2-(-x+3)^2=2-(3-x)^2=R(x).\] Replacing \(x\) by \(-x\) reflects the graph in the y-axis.

The point mapping is (x,y) to (-x,y).

The maximum is \(5\) and the minimum is \(1\), so \[a=\frac{5-1}{2}=2,\qquad c=\frac{5+1}{2}=3.\]

The positive amplitude is half the range; the vertical shift is the midline.

A maximum occurs at \(x=-60^\circ\). For positive \(a\), cosine is maximal when \(x+b=0^\circ\) modulo \(360^\circ\), giving \(b=60\) in the stated interval. Hence \(y=2\cos(x+60)^\circ+3\).

Read the phase from a maximum and apply 0<b<360.

The turning point of \(y=f(x)\) is \((-6,-4)\).

For

the graph moves 5 units up:

For

the \(x\)-coordinate is divided by 3:

For the sketch of \(y=2g(x)\), keep every \(x\)-coordinate the same and double every \(y\)-coordinate.

Draw straight line segments through these points for \(-1\le x\le 7\).

For part (c), \(h(x)\) has roots at \(x=-1\) and \(x=6\).

For \(h(x+a)\) to pass through \((2,0)\),

This is in completed square form, so the maximum point is

For the translation

the graph moves 4 units right. Replace \(x\) by \(x-4\):

The curve \(T\) has equation

So \(C\) is reflected in the \(x\)-axis.

For the cosine graph, the maximum is 5 and the minimum is \(-1\).

For

a maximum occurs when \(x-b=0\). The graph has a maximum at \(x=270^\circ\), so

The maximum point of \(y=f(x)\) is \((3,5)\).

For

the \(y\)-coordinate is multiplied by 2:

For

the graph moves 7 units down:

For

the graph reflects in the \(y\)-axis:

For part (b),

So the graph moves 2 units right.

Replace \(x\) by \(x-2\):