Graph Transformations: Year 11 Mathematics Advanced Guide

Take a point on a graph, such as \((2,3)\). Now replace every \(x\) in the equation with \(x-4\). Which way does the point move?

You might expect it to move four units left because of the minus sign. In fact, it moves four units right, to \((6,3)\).

This is the central challenge of graph transformations: the changes you make to an equation do not always have the direction or effect you first expect. Once you understand how individual points move, however, reflections, translations, and dilations become much easier to predict.

This guide covers the graph transformations and circle equations in the NSW 2024 Year 11 Mathematics Advanced syllabus. All function examples use non-trigonometric functions, in line with the scope of this focus area.

01Start by tracking a point

Consider a function \(y=f(x)\), with a point \(P=(u,v)\) on its graph.

This means that:

\[ v=f(u) \]

A transformation changes the position of this point. The most reliable way to understand any transformation is to work out where \((u,v)\) moves.

For example, suppose we know that \((2,3)\) lies on a graph. If the transformed graph contains the point \((6,3)\), we can see that the \(x\)-coordinate has increased by four, while the \(y\)-coordinate has stayed the same.

The graph has moved horizontally, not vertically.

This point-tracking approach gives us a consistent way to understand every transformation in this topic.

Investigating transformations with graphing applications

The NSW 2024 syllabus expects you to establish the effects of transformations using graphing applications.

You can investigate these effects using Desmos, GeoGebra, or another graphing application.

Start by entering:

\[ y=x+1 \]

Choose the point \((1,2)\), which lies on this graph. Now enter the following equations on the same set of axes.

Each equation represents a different transformation of the original graph.

Notice that the first two equations produce reflections, the next two produce translations, and the final two produce dilations.

The important observation is that the coordinates change in predictable ways. We can now build rules from those observations.

02Reflections: flipping a graph across an axis

Imagine holding a sheet of transparent paper with a graph drawn on it.

If you fold the paper along the \(y\)-axis, every point on the right moves to the corresponding position on the left. The vertical coordinate stays the same.

This is a reflection in the \(y\)-axis.

If you fold the paper along the \(x\)-axis, the horizontal coordinate stays the same, while the vertical coordinate changes sign.

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

Reflection in the y-axis

To reflect a graph in the \(y\)-axis, replace \(x\) with \(-x\) in its equation.

If the original function is:

\[ y=f(x) \]

the reflected function is:

\[ \boxed{y=f(-x)} \]

The point mapping is:

\[ (u,v)\longrightarrow(-u,v) \]

The \(x\)-coordinate changes sign, but the \(y\)-coordinate stays unchanged.

For example, consider:

\[ y=x+2 \]

Reflecting in the \(y\)-axis gives:

\[ y=-x+2 \]

The original point \((1,3)\) becomes \((-1,3)\).

Notice that the reflected graph is not obtained by simply putting a negative sign in front of the entire function.

Reflection in the x-axis

To reflect a graph in the \(x\)-axis, replace \(y\) with \(-y\) in the original equation.

Starting with:

\[ y=f(x) \]

we obtain:

\[ -y=f(x) \]

Rearranging gives:

\[ \boxed{y=-f(x)} \]

The point mapping is:

\[ (u,v)\longrightarrow(u,-v) \]

The \(y\)-coordinate changes sign, while the \(x\)-coordinate stays unchanged.

For example:

\[ y=x+2 \]

becomes:

\[ y=-x-2 \]

The point \((1,3)\) becomes \((1,-3)\).

A common mistake: confusing f(-x) with -f(x)

These two expressions produce different transformations.

A useful way to remember the difference:

• A negative sign inside the function changes the horizontal position.
• A negative sign outside the function changes the vertical position.

There is one complication worth knowing. If the original graph is symmetrical about an axis, reflecting it in that axis might not visibly change the graph.

For example, \(y=x^2\) is already symmetrical about the \(y\)-axis. Therefore:

\[ f(-x)=(-x)^2=x^2 \]

The reflected graph looks identical to the original.

This does not mean the transformation failed. It means the graph already had the required symmetry.

Reflecting in both axes

If you reflect a graph in both axes, the point mapping becomes:

\[ (u,v)\longrightarrow(-u,-v) \]

This is equivalent to a rotation of \(180^\circ\) about the origin.

For a function:

\[ y=f(x) \]

the resulting graph is:

\[ y=-f(-x) \]

This can be useful when a question describes several transformations rather than giving you the final equation directly.

03Translations: moving a graph without changing its shape

A translation moves every point on a graph by the same distance and in the same direction.

Imagine sliding a transparent sheet across your desk. The graph keeps exactly the same shape, orientation, and size. Only its position changes.

There are two basic translations:

• Horizontal translations move a graph left or right.
• Vertical translations move a graph up or down.

Horizontal translations

Suppose the original function is:

\[ y=f(x) \]

Replacing \(x\) with \(x-a\) gives:

\[ \boxed{y=f(x-a)} \]

The corresponding point mapping is:

\[ (u,v)\longrightarrow(u+a,v) \]

Therefore:

• If \(a>0\), the graph moves right by \(a\) units.
• If \(a<0\), the graph moves left by \(|a|\) units.

For example:

\[ y=\sqrt{x} \]

becomes:

\[ y=\sqrt{x-3} \]

The graph moves three units to the right.

The original endpoint \((0,0)\) becomes \((3,0)\).

The new domain is \(x\geq3\), rather than \(x\geq0\).

The sign inside the brackets is the source of many mistakes.

Compare:

\[ y=f(x-3) \]

and:

\[ y=f(x+3) \]

The first moves right three units. The second moves left three units.

The sign is opposite to the direction of movement.

Vertical translations

To translate a graph vertically, replace \(y\) with \(y-b\) in its original equation.

For a function, this gives:

\[ y-b=f(x) \]

Rearranging:

\[ \boxed{y=f(x)+b} \]

The point mapping is:

\[ (u,v)\longrightarrow(u,v+b) \]

Therefore:

• If \(b>0\), the graph moves up by \(b\) units.
• If \(b<0\), the graph moves down by \(|b|\) units.

For example:

\[ y=x^2 \]

becomes:

\[ y=x^2+4 \]

The graph moves four units upwards.

Its vertex changes from \((0,0)\) to \((0,4)\).

Similarly:

\[ y=x^2-2 \]

moves the original graph two units downwards.

Combining horizontal and vertical translations

The two translations can be combined into one function rule:

\[ \boxed{y=f(x-a)+b} \]

The original point mapping becomes:

\[ \boxed{(u,v)\longrightarrow(u+a,v+b)} \]

For example, consider:

\[ y=(x-2)^2+5 \]

Starting with \(y=x^2\):

• The \(x-2\) translates the graph two units right.
• The \(+5\) translates the graph five units up.

The vertex moves from \((0,0)\) to \((2,5)\).

The shape of the parabola does not change.

This is the foundation for understanding more complicated combinations of transformations.

04Dilations: stretching and compressing a graph

A translation moves a graph without changing its size.

A dilation changes the distance of points from an axis.

There are two types of dilation:

• A horizontal dilation stretches or compresses the graph from the \(y\)-axis.
• A vertical dilation stretches or compresses the graph from the \(x\)-axis.

The phrase from the axis is important. It tells you which coordinate changes.

Horizontal dilation

Consider a graph with an original point:

\[ P=(u,v) \]

A horizontal dilation by a factor of \(k\) changes the point to:

\[ \boxed{(u,v)\longrightarrow(ku,v)} \]

The horizontal coordinate is multiplied by \(k\). The vertical coordinate stays the same.

To produce this transformation algebraically, replace \(x\) with \(\frac{x}{k}\) in the original function:

\[ \boxed{y=f\left(\frac{x}{k}\right)} \]

For \(k>1\), the graph stretches horizontally.

For example:

\[ y=f\left(\frac{x}{2}\right) \]

produces a horizontal dilation by a factor of 2.

Every point moves twice as far from the \(y\)-axis.

If the original point was \((1,3)\), its new position is \((2,3)\).

What happens when the dilation factor is between 0 and 1?

This is where you need to be particularly careful.

Suppose:

\[ k=\frac12 \]

The transformed function is:

\[ y=f\left(\frac{x}{1/2}\right) \]

which simplifies to:

\[ y=f(2x) \]

The point mapping is:

\[ (u,v)\longrightarrow\left(\frac{u}{2},v\right) \]

Every point moves halfway towards the \(y\)-axis.

The graph is compressed horizontally by a factor of \(\frac12\).

In general:

A smaller positive factor brings the graph closer to the \(y\)-axis.

Vertical dilation

A vertical dilation changes the distance of every point from the \(x\)-axis.

The point mapping is:

\[ \boxed{(u,v)\longrightarrow(u,lv)} \]

where \(l\) is the vertical dilation factor.

To obtain this transformation, replace \(y\) with \(\frac{y}{l}\) in the original graph equation.

For an explicit function, the resulting rule is:

\[ \boxed{y=lf(x)} \]

For \(l>1\), the graph stretches vertically.

For example:

\[ y=2x^2 \]

is a vertical dilation of \(y=x^2\) by a factor of 2.

The point \((1,1)\) moves to \((1,2)\).

The horizontal coordinate stays the same, but the vertical coordinate doubles.

What happens when the vertical factor is between 0 and 1?

Suppose:

\[ l=\frac12 \]

The transformation becomes:

\[ y=\frac12 f(x) \]

The point mapping is:

\[ (u,v)\longrightarrow\left(u,\frac{v}{2}\right) \]

Every point moves halfway towards the \(x\)-axis.

The graph is vertically compressed by a factor of \(\frac12\).

The general rules are:

Horizontal versus vertical dilation

These two transformations are easy to confuse because both change the appearance of a graph.

However, they act in different directions.

For example, if the original graph is:

\[ y=x^2 \]

a horizontal dilation by 2 gives:

\[ y=\left(\frac{x}{2}\right)^2 =\frac{x^2}{4} \]

A vertical dilation by 2 gives:

\[ y=2x^2 \]

These are not the same graph.

The first becomes wider. The second becomes steeper and narrower.

Uniform dilation: changing both coordinates

There is one more type of dilation in the syllabus.

Suppose both coordinates are multiplied by the same positive factor \(k\).

The point mapping becomes:

\[ \boxed{(u,v)\longrightarrow(ku,kv)} \]

This is a dilation about the origin.

To obtain it from a graph equation, replace both \(x\) and \(y\) with their corresponding divided values:

\[ x\longrightarrow\frac{x}{k} \]

\[ y\longrightarrow\frac{y}{k} \]

For a general graph equation \(F(x,y)=0\), the transformed equation is:

\[ \boxed{F\left(\frac{x}{k},\frac{y}{k}\right)=0} \]

For a function \(y=f(x)\), this can be written as:

\[ \boxed{y=kf\left(\frac{x}{k}\right)} \]

A uniform dilation preserves the shape of the graph. It changes its overall size without stretching it more in one direction than another.

For example, consider a circle centred at the origin:

\[ x^2+y^2=4 \]

Its radius is 2.

After a uniform dilation by a factor of 3:

\[ \frac{x^2}{9}+\frac{y^2}{9}=4 \]

Multiplying through by 9:

\[ x^2+y^2=36 \]

The new radius is 6.

The circle has enlarged by a factor of 3.

The effect depends on the factor:

• If \(k>1\), the graph enlarges.
• If \(0<k<1\), the graph reduces in size.
• If \(k=1\), the graph is unchanged.

A dilation factor of zero is not allowed because it collapses the graph onto an axis or a point rather than producing a standard dilation.

If a signed factor is negative, it also introduces a reflection. For example, a horizontal factor of \(-2\) combines a horizontal stretch by 2 with a reflection in the \(y\)-axis.

For most dilation questions, the stated factor will be positive. The distinction is useful when interpreting a more complicated equation.

05Combining transformations into one function rule

So far, we have considered each transformation separately.

In an exam, however, you will often be given a function that contains several transformations at once.

A useful general form is:

\[ \boxed{g(x)=lf\left(\frac{x-a}{k}\right)+b} \]

Here:

• \(a\) determines the horizontal translation.
• \(b\) determines the vertical translation.
• \(k\) determines the horizontal dilation.
• \(l\) determines the vertical dilation and may also introduce a reflection if negative.

For positive \(k\) and \(l\), the point mapping is:

\[ \boxed{(u,v)\longrightarrow(a+ku,b+lv)} \]

This single formula is one of the most useful tools in graph transformation questions.

It tells you exactly where every point on the original graph moves.

Reading the transformations from the equation

Consider:

\[ g(x)=3f\left(\frac{x-4}{2}\right)-5 \]

Compare this with:

\[ g(x)=lf\left(\frac{x-a}{k}\right)+b \]

We can identify:

\[ a=4,\quad b=-5,\quad k=2,\quad l=3 \]

Therefore, the transformations are:

• Horizontal dilation by a factor of 2.
• Vertical dilation by a factor of 3.
• Translation four units right.
• Translation five units down.

The point mapping is:

\[ (u,v)\longrightarrow(4+2u,-5+3v) \]

For example, if \((1,2)\) lies on the original graph, its transformed position is:

\[ (1,2)\longrightarrow(6,1) \]

because:

\[ x=4+2(1)=6 \]

and:

\[ y=-5+3(2)=1 \]

You can use this method for almost any combination of translations and dilations.

If \(k\) or \(l\) is negative, the same coordinate mapping still works, but the corresponding coordinate also changes sign. This introduces the relevant reflection.

The domain and range of a transformed function

A transformation changes more than the appearance of a graph. It can also change its domain and range.

Suppose the original function has:

\[ \text{Domain}=D \]

\[ \text{Range}=R \]

For the transformed function:

\[ g(x)=lf\left(\frac{x-a}{k}\right)+b \]

the new domain is:

\[ \boxed{D_g=\{a+ku:u\in D\}} \]

The new range is:

\[ \boxed{R_g=\{b+lv:v\in R\}} \]

These expressions describe how every allowed input and output changes.

For positive \(k\) and \(l\), the rules are particularly straightforward.

If the original domain is:

\[ [p,q] \]

then the transformed domain is:

\[ [a+kp,a+kq] \]

If the original range is:

\[ [r,s] \]

then the transformed range is:

\[ [b+lr,b+ls] \]

For negative factors, check the order of the endpoints because the direction is reversed.

You must also preserve excluded points. If an original domain excludes a particular input, its transformed position remains excluded.

This matters when working with reciprocal functions, removable discontinuities, and restricted domains.

06Intercepts, asymptotes, and discontinuities after transformations

Knowing how to identify the transformations is only part of the job.

Questions can also ask you to determine the important features of the transformed graph.

These include:

• \(x\)-intercepts and \(y\)-intercepts.
• Vertical and horizontal asymptotes.
• Removable discontinuities, such as holes.
• Domain and range.

The point mapping gives us a systematic way to handle all of them.

Finding intercepts

An \(x\)-intercept occurs where:

\[ y=0 \]

A \(y\)-intercept occurs where:

\[ x=0 \]

For a transformed function:

\[ g(x)=lf\left(\frac{x-a}{k}\right)+b \]

the \(x\)-intercepts satisfy:

\[ lf\left(\frac{x-a}{k}\right)+b=0 \]

Rearranging:

\[ f\left(\frac{x-a}{k}\right)=-\frac{b}{l} \]

This means you can find the relevant points on the original graph first, then transform them.

For the \(y\)-intercept, substitute \(x=0\), provided that zero belongs to the transformed domain.

Finding asymptotes

An asymptote is a line that a graph approaches as the input or output changes.

Transformations move asymptotes along with the rest of the graph.

Suppose the original graph has a vertical asymptote:

\[ x=c \]

After the transformation:

\[ x_{\text{new}}=a+kc \]

Therefore, the new vertical asymptote is:

\[ \boxed{x=a+kc} \]

Similarly, if the original graph has a horizontal asymptote:

\[ y=d \]

the transformed horizontal asymptote is:

\[ \boxed{y=b+ld} \]

These rules follow directly from the coordinate mapping.

For example, if a graph has a vertical asymptote at \(x=1\), and you apply a horizontal dilation by 2 followed by a translation three units right, the new asymptote is:

\[ x=3+2(1)=5 \]

It is now located at \(x=5\).

Finding discontinuities and holes

A removable discontinuity occurs when a graph has a missing point, even though the surrounding graph approaches a definite value.

Imagine a straight line with one small gap in it.

A transformation moves the gap just like any other point.

If the original hole is at:

\[ (c,d) \]

the transformed hole is at:

\[ \boxed{(a+kc,b+ld)} \]

The hole must remain excluded from the transformed graph.

However, there is an important distinction between a missing point and a missing output value.

If another point on the graph has the same \(y\)-coordinate, that output value may still belong to the range.

This distinction becomes important in the final practice question.

07Circles: translations, centres, and radii

The transformation principles also apply to circles.

A circle centred at the origin with radius \(r\) has the equation:

\[ x^2+y^2=r^2 \]

Every point on the circle is exactly \(r\) units from the origin.

Now imagine translating the entire circle.

Its shape and radius remain unchanged, but its centre moves.

This gives the general equation of a translated circle:

\[ \boxed{(x-a)^2+(y-b)^2=r^2} \]

Here:

• \(a\) is the horizontal coordinate of the centre.
• \(b\) is the vertical coordinate of the centre.
• \(r\) is the radius.

Therefore, the centre is:

\[ \boxed{(a,b)} \]

and the radius is:

\[ \boxed{r} \]

We normally take \(r>0\). More generally, the geometric radius is \(|r|\), since the equation contains \(r^2\).

Why the signs in the equation look reversed

Compare the general equation:

\[ (x-a)^2+(y-b)^2=r^2 \]

with the original circle:

\[ x^2+y^2=r^2 \]

The replacement \(x\to x-a\) translates the graph horizontally by \(a\).

The replacement \(y\to y-b\) translates the graph vertically by \(b\).

Therefore:

• \(x-4\) places the centre four units right.
• \(x+4\) places the centre four units left.
• \(y-3\) places the centre three units up.
• \(y+3\) places the centre three units down.

For example:

\[ (x-2)^2+(y+3)^2=25 \]

has centre:

\[ \boxed{(2,-3)} \]

and radius:

\[ \boxed{5} \]

The circle has moved two units right and three units down from a circle of radius 5 centred at the origin.

Finding the domain and range of a circle

A circle is not a function because many vertical lines intersect it twice.

However, we can still identify its domain and range.

For a circle with centre \((a,b)\) and radius \(r\):

The leftmost point has \(x\)-coordinate \(a-r\).

The rightmost point has \(x\)-coordinate \(a+r\).

Therefore:

\[ \boxed{\text{Domain}=[a-r,a+r]} \]

Similarly, the lowest point has \(y\)-coordinate \(b-r\), and the highest point has \(y\)-coordinate \(b+r\).

Therefore:

\[ \boxed{\text{Range}=[b-r,b+r]} \]

For example, consider:

\[ (x-2)^2+(y+3)^2=25 \]

We already know that the centre is \((2,-3)\), and the radius is 5.

The domain is:

\[ [2-5,2+5] \]

\[ \boxed{[-3,7]} \]

The range is:

\[ [-3-5,-3+5] \]

\[ \boxed{[-8,2]} \]

The endpoints are included because the circle itself contains its leftmost, rightmost, highest, and lowest points.

Graphing a circle from its equation

When you are asked to sketch a circle, follow this process.

Step 1: Identify the centre.

Read the values of \(a\) and \(b\) from the equation.

Step 2: Identify the radius.

Take the positive square root of the right-hand side.

Step 3: Plot the centre.

Mark \((a,b)\) on your coordinate axes.

Step 4: Find the four extreme points.

Move \(r\) units left, right, up, and down from the centre.

These points are:

\[ (a-r,b),\quad(a+r,b) \]

\[ (a,b-r),\quad(a,b+r) \]

Step 5: Draw the circle.

Join the four points with a smooth curve, maintaining equal distance from the centre.

08Completing the square to find a circle’s centre and radius

Not every circle equation will be given in the convenient form:

\[ (x-a)^2+(y-b)^2=r^2 \]

Sometimes you will receive an expanded equation:

\[ \boxed{x^2+y^2+Ax+By+C=0} \]

Here, \(A\), \(B\), and \(C\) are constants.

Your job is to rearrange the equation into the standard circle form.

The technique is called completing the square.

It is the same algebraic process you may have used when rewriting a quadratic expression into vertex form.

Understanding the method

Consider:

\[ x^2+6x \]

To complete the square, take half the coefficient of \(x\), then square it.

Half of 6 is 3.

Squaring gives:

\[ 3^2=9 \]

Therefore:

\[ x^2+6x+9=(x+3)^2 \]

The same principle applies to the \(y\)-terms.

For a general expression:

\[ x^2+Ax \]

we obtain:

\[ x^2+Ax+\left(\frac A2\right)^2 = \left(x+\frac A2\right)^2 \]

For:

\[ y^2+By \]

we obtain:

\[ y^2+By+\left(\frac B2\right)^2 = \left(y+\frac B2\right)^2 \]

The key is to add the correct square terms without changing the equality.

The general formula

Starting with:

\[ x^2+y^2+Ax+By+C=0 \]

Group the \(x\)-terms and \(y\)-terms:

\[ (x^2+Ax)+(y^2+By)=-C \]

Complete both squares:

\[ \left(x+\frac A2\right)^2 + \left(y+\frac B2\right)^2 = \frac{A^2}{4}+\frac{B^2}{4}-C \]

Therefore:

\[ \boxed{ \left(x+\frac A2\right)^2+ \left(y+\frac B2\right)^2 = \frac{A^2+B^2}{4}-C } \]

Comparing this with the standard circle equation gives:

\[ \boxed{\text{Centre}= \left(-\frac A2,-\frac B2\right)} \]

and:

\[ \boxed{ r=\sqrt{\frac{A^2+B^2}{4}-C} } \]

The quantity inside the square root must be positive for the equation to represent a circle with a positive radius.

If it equals zero, the graph is a single point.

If it is negative, there are no real points on the graph.

09Finding the equation of a circle from its graph

Sometimes you will be given a graph rather than an equation.

Your task is to extract the centre and radius from the graph, then construct the equation.

This is the reverse of the process we have just used.

Remember:

\[ (x-a)^2+(y-b)^2=r^2 \]

You need only three pieces of information:

• The horizontal coordinate of the centre.
• The vertical coordinate of the centre.
• The radius.

Finding the centre from extreme points

The topmost and bottommost points of a circle lie directly above and below its centre.

Likewise, the leftmost and rightmost points lie directly to the left and right of its centre.

This gives you a useful way to recover the centre even when it is not marked on the graph.

For example, suppose a circle has:

• A highest point at \((2,5)\).
• A leftmost point at \((-2,1)\).

The highest point shares its \(x\)-coordinate with the centre.

Therefore, the centre has:

\[ a=2 \]

The leftmost point shares its \(y\)-coordinate with the centre.

Therefore:

\[ b=1 \]

The centre is:

\[ \boxed{(2,1)} \]

The vertical distance from the centre to the highest point is:

\[ r=5-1=4 \]

Therefore, the radius is 4.

The equation is:

\[ \boxed{(x-2)^2+(y-1)^2=16} \]

The domain is:

\[ [-2,6] \]

The range is:

\[ [-3,5] \]

10The order of transformations matters

There is one final idea that brings the entire topic together.

Two transformations applied to the same graph do not necessarily produce the same result when you reverse their order.

This is particularly important when you combine horizontal translations and horizontal dilations.

Consider the original graph:

\[ y=x^2 \]

Its vertex is at \((0,0)\).

We will apply two transformations:

• Translate the graph one unit to the right.
• Apply a horizontal dilation by a factor of 2.

There are two possible orders.

Order A: Translate first, then dilate

First, translate the graph one unit right:

\[ y=(x-1)^2 \]

The vertex is now:

\[ (1,0) \]

Next, apply a horizontal dilation by a factor of 2.

Replace \(x\) with \(\frac{x}{2}\):

\[ y=\left(\frac{x}{2}-1\right)^2 \]

Simplifying:

\[ \boxed{y=\frac{(x-2)^2}{4}} \]

The vertex is now:

\[ \boxed{(2,0)} \]

The translation has also been stretched because it was applied before the dilation.

Order B: Dilate first, then translate

Start again with:

\[ y=x^2 \]

First, apply the horizontal dilation by a factor of 2:

\[ y=\left(\frac{x}{2}\right)^2 \]

The vertex remains at:

\[ (0,0) \]

Now translate the graph one unit right.

Replace \(x\) with \(x-1\):

\[ y=\left(\frac{x-1}{2}\right)^2 \]

Therefore:

\[ \boxed{y=\frac{(x-1)^2}{4}} \]

The vertex is:

\[ \boxed{(1,0)} \]

Comparing the results

The two final equations are:

\[ y=\frac{(x-2)^2}{4} \]

and:

\[ y=\frac{(x-1)^2}{4} \]

They are different.

The first graph has vertex \((2,0)\), while the second has vertex \((1,0)\).

Therefore:

\[ \boxed{\text{The order of these transformations matters.}} \]

Why does the order matter?

A horizontal dilation changes every horizontal coordinate.

If you translate a graph first, the translation is included in the coordinates being stretched.

If you dilate first, the translation happens afterwards and is not stretched.

The same principle applies to vertical transformations.

For example, a vertical stretch followed by a vertical translation does not generally produce the same graph as a vertical translation followed by a vertical stretch.

However, transformations affecting different coordinates can commute in these simple cases. A horizontal dilation and a vertical translation, for instance, produce the same result whichever is applied first.

The important decision is to track the coordinates that each transformation changes.

A reliable method for complicated transformations

When a question gives you a sequence of transformations:

1. Write down the original function or graph equation.
2. Apply the first transformation to the current equation.
3. Use the resulting equation for the next transformation.
4. Continue until all transformations have been applied.
5. Check the final graph using an important point, such as its vertex, endpoint, or asymptote.

Do not combine all the signs mentally and hope they work out.

Writing each step separately is usually faster than correcting an incorrect final equation.

11Common mistakes to avoid

Before moving to the practice questions, make sure you can recognise these mistakes.

The most reliable habit is to track an original point through every transformation. It removes much of the guesswork.

12Questions and solutions

These questions progress from standard Year 11 applications to more demanding problems involving several transformations at once.

Try each question before reading its solution. In particular, pay attention to the restrictions on domains, the order of transformations, and the difference between a missing point and a missing range value.

Question 1

Consider the function:

\[ y=-(x+3)^2+4 \]

Starting with the graph \(y=x^2\):

(a) Describe the transformations applied to the original graph.

(b) Find the coordinates of the vertex.

(c) Find the \(x\)-intercepts and \(y\)-intercept.

(d) State the domain and range.

Solution 1

The graph is a parabola reflected in the \(x\)-axis, translated three units left, and translated four units up.

(a) Transformations

The function is:

\[ y=-(x+3)^2+4 \]

The expression \(x+3\) translates the graph three units left.

The negative sign outside the squared expression reflects the graph in the \(x\)-axis.

The \(+4\) translates the graph four units upwards.

(b) Vertex

The original vertex is \((0,0)\).

After the horizontal and vertical translations, it becomes:

\[ \boxed{(-3,4)} \]

The reflection does not change the vertex’s horizontal position.

(c) Intercepts

For the \(x\)-intercepts, set \(y=0\):

\[ -(x+3)^2+4=0 \]

\[ (x+3)^2=4 \]

Taking square roots:

\[ x+3=\pm2 \]

Therefore:

\[ x=-3\pm2 \]

The \(x\)-intercepts are:

\[ \boxed{(-5,0)\text{ and }(-1,0)} \]

For the \(y\)-intercept, substitute \(x=0\):

\[ y=-(0+3)^2+4 \]

\[ y=-9+4 \]

\[ \boxed{y=-5} \]

The \(y\)-intercept is \((0,-5)\).

(d) Domain and range

A quadratic function is defined for every real value of \(x\).

Therefore:

\[ \boxed{\text{Domain}=\mathbb R} \]

The parabola opens downwards because of the negative coefficient.

Its maximum value is 4, occurring at the vertex.

Therefore:

\[ \boxed{\text{Range}=(-\infty,4]} \]

The graph has a maximum, not a minimum. This is a useful check when deciding the range of a reflected parabola.

Question 2

Consider the circle:

\[ x^2+y^2-4x+6y-12=0 \]

(a) Complete the square to express the equation in standard circle form.

(b) Find the centre and radius.

(c) Determine the domain and range.

(d) Find all \(x\)-intercepts and \(y\)-intercepts.

Solution 2

The circle has centre \((2,-3)\) and radius 5, with domain \([-3,7]\) and range \([-8,2]\).

(a) Complete the square

Start with:

\[ x^2+y^2-4x+6y-12=0 \]

Group the variables and move the constant:

\[ (x^2-4x)+(y^2+6y)=12 \]

For the \(x\)-terms, half of \(-4\) is \(-2\), and its square is 4.

For the \(y\)-terms, half of 6 is 3, and its square is 9.

Add both values to both sides:

\[ (x-2)^2+(y+3)^2=12+4+9 \]

Therefore:

\[ \boxed{(x-2)^2+(y+3)^2=25} \]

(b) Centre and radius

Compare with:

\[ (x-a)^2+(y-b)^2=r^2 \]

Therefore:

\[ \boxed{\text{Centre}=(2,-3)} \]

and:

\[ \boxed{r=5} \]

(c) Domain and range

The domain extends 5 units either side of the centre’s \(x\)-coordinate:

\[ 2-5\leq x\leq2+5 \]

Therefore:

\[ \boxed{\text{Domain}=[-3,7]} \]

The range extends 5 units either side of the centre’s \(y\)-coordinate:

\[ -3-5\leq y\leq-3+5 \]

Therefore:

\[ \boxed{\text{Range}=[-8,2]} \]

(d) Intercepts

For the \(x\)-intercepts, set \(y=0\):

\[ (x-2)^2+(0+3)^2=25 \]

\[ (x-2)^2+9=25 \]

\[ (x-2)^2=16 \]

Therefore:

\[ x-2=\pm4 \]

\[ \boxed{x=-2\text{ or }x=6} \]

For the \(y\)-intercepts, set \(x=0\):

\[ (0-2)^2+(y+3)^2=25 \]

\[ 4+(y+3)^2=25 \]

\[ (y+3)^2=21 \]

Therefore:

\[ \boxed{y=-3\pm\sqrt{21}} \]

The circle has two intercepts on each coordinate axis.

Question 3

Consider the original function:

\[ f(x)=\frac{1}{x-1} \]

where \(x\neq1\).

A transformed function is defined by:

\[ g(x)=-2f\left(\frac{x+2}{3}\right)+1 \]

(a) Describe the transformations applied to the original graph.

(b) Find the simplified rule for \(g(x)\).

(c) Determine its domain and range.

(d) Find the vertical and horizontal asymptotes.

(e) Find both intercepts.

Solution 3

The transformed function is:

\[ \boxed{g(x)=-\frac{6}{x-1}+1} \]

Its domain excludes \(x=1\), its range excludes \(y=1\), and its asymptotes are \(x=1\) and \(y=1\).

(a) Transformations

Compare:

\[ g(x)=lf\left(\frac{x-a}{k}\right)+b \]

with:

\[ g(x)=-2f\left(\frac{x+2}{3}\right)+1 \]

We identify:

\[ a=-2,\quad b=1,\quad k=3,\quad l=-2 \]

Therefore, the graph undergoes:

• A horizontal dilation by a factor of 3.
• A vertical dilation by a factor of 2.
• A reflection in the \(x\)-axis.
• A translation two units left.
• A translation one unit up.

The point mapping is:

\[ (u,v)\longrightarrow(-2+3u,1-2v) \]

(b) Simplified function rule

Substitute the original function:

\[ g(x)=-2\left(\frac{1}{\frac{x+2}{3}-1}\right)+1 \]

Simplify the denominator:

\[ \frac{x+2}{3}-1 = \frac{x-1}{3} \]

Therefore:

\[ g(x)=-2\left(\frac{3}{x-1}\right)+1 \]

Hence:

\[ \boxed{g(x)=-\frac{6}{x-1}+1} \]

(c) Domain and range

The original function is undefined when:

\[ x-1=0 \]

Therefore, its domain excludes \(x=1\).

The transformed horizontal coordinate of this excluded input is:

\[ x=-2+3(1)=1 \]

Therefore:

\[ \boxed{\text{Domain}=\mathbb R\setminus\{1\}} \]

The original reciprocal function never produces an output of zero.

After the vertical transformations, that excluded output becomes:

\[ y=1-2(0)=1 \]

Therefore:

\[ \boxed{\text{Range}=\mathbb R\setminus\{1\}} \]

(d) Asymptotes

The original vertical asymptote is \(x=1\).

It transforms to:

\[ x=-2+3(1)=1 \]

The original horizontal asymptote is \(y=0\).

It transforms to:

\[ y=1-2(0)=1 \]

Therefore:

\[ \boxed{x=1,\qquad y=1} \]

(e) Intercepts

For the \(x\)-intercept:

\[ -\frac{6}{x-1}+1=0 \]

\[ \frac{6}{x-1}=1 \]

Therefore:

\[ \boxed{x=7} \]

The \(x\)-intercept is \((7,0)\).

For the \(y\)-intercept, substitute \(x=0\):

\[ g(0)=-\frac{6}{-1}+1 \]

\[ g(0)=6+1 \]

\[ \boxed{g(0)=7} \]

The \(y\)-intercept is \((0,7)\).

A useful final check is that both intercepts lie in the domain and that neither asymptote is crossed by the graph.

Question 4

A circle is tangent to the \(y\)-axis at the point:

\[ T=(0,3) \]

It also passes through:

\[ P=(4,0) \]

Find:

(a) The centre of the circle.

(b) Its radius and equation.

(c) Its domain and range.

(d) Both \(x\)-intercepts.

Solution 4

The circle has centre \(\left(\frac{25}{8},3\right)\), radius \(\frac{25}{8}\), and equation:

\[ \boxed{ \left(x-\frac{25}{8}\right)^2+(y-3)^2 = \frac{625}{64} } \]

(a) Centre

The key information is that the circle is tangent to the \(y\)-axis at \((0,3)\).

A radius drawn to the point of tangency must be perpendicular to the tangent line.

Since the \(y\)-axis is vertical, the radius must be horizontal.

Therefore, the centre must lie on the horizontal line:

\[ y=3 \]

Let the centre be:

\[ C=(h,3) \]

The radius is the horizontal distance from the centre to the \(y\)-axis:

\[ r=|h| \]

(b) Radius and equation

The point \(P=(4,0)\) lies on the circle.

Its distance from the centre must equal the radius.

Using the distance formula:

\[ (4-h)^2+(0-3)^2=h^2 \]

Expand:

\[ (4-h)^2+9=h^2 \]

\[ 16-8h+h^2+9=h^2 \]

Cancel \(h^2\) from both sides:

\[ 25-8h=0 \]

Therefore:

\[ 8h=25 \]

\[ \boxed{h=\frac{25}{8}} \]

The centre is:

\[ \boxed{C=\left(\frac{25}{8},3\right)} \]

The radius is:

\[ \boxed{r=\frac{25}{8}} \]

Substitute into the standard circle equation:

\[ (x-a)^2+(y-b)^2=r^2 \]

We obtain:

\[ \boxed{ \left(x-\frac{25}{8}\right)^2+(y-3)^2 = \frac{625}{64} } \]

(c) Domain and range

The domain extends from the leftmost point to the rightmost point.

\[ x_{\min}=\frac{25}{8}-\frac{25}{8}=0 \]

\[ x_{\max}=\frac{25}{8}+\frac{25}{8} =\frac{25}{4} \]

Therefore:

\[ \boxed{\text{Domain}=\left[0,\frac{25}{4}\right]} \]

The range extends from the lowest point to the highest point:

\[ y_{\min}=3-\frac{25}{8} \]

\[ =\frac{24}{8}-\frac{25}{8} =-\frac18 \]

And:

\[ y_{\max}=3+\frac{25}{8} \]

\[ =\frac{24}{8}+\frac{25}{8} =\frac{49}{8} \]

Therefore:

\[ \boxed{\text{Range}=\left[-\frac18,\frac{49}{8}\right]} \]

(d) \(x\)-intercepts

Set \(y=0\):

\[ \left(x-\frac{25}{8}\right)^2+(0-3)^2 = \frac{625}{64} \]

\[ \left(x-\frac{25}{8}\right)^2+9 = \frac{625}{64} \]

Since \(9=\frac{576}{64}\):

\[ \left(x-\frac{25}{8}\right)^2 = \frac{49}{64} \]

Taking square roots:

\[ x-\frac{25}{8}=\pm\frac78 \]

Therefore:

\[ x=\frac{25\pm7}{8} \]

Giving:

\[ \boxed{x=4\quad\text{or}\quad x=\frac94} \]

The two \(x\)-intercepts are:

\[ \boxed{(4,0)\text{ and }\left(\frac94,0\right)} \]

The given point \(P=(4,0)\) is one of these intercepts. The second intercept lies further along the positive \(x\)-axis towards the point of tangency.

The key geometric idea was that a radius is perpendicular to a tangent. Without using that relationship, the centre would be much harder to locate.

Question 5

Consider the graph of:

\[ y=x^2 \]

with the restricted domain:

\[ -2\leq x\leq3,\qquad x\neq1 \]

This means that the graph includes both endpoints, but has a hole at \((1,1)\).

Two students apply the same transformations in different orders.

Student A:

1. Translates the original graph one unit to the right.
2. Applies a horizontal dilation by a factor of 2 about the \(y\)-axis.

Student B:

1. Applies a horizontal dilation by a factor of 2 about the \(y\)-axis.
2. Translates the original graph one unit to the right.

Answer the following.

(a) Find the final function rule for each student.

(b) Determine the domain and range of each transformed graph.

(c) Find the position of the hole in each graph.

(d) Find the vertex of each graph.

(e) Determine whether the two transformed graphs intersect. If they do, find all intersection points.

(f) A student argues that because both graphs have a hole at a point with \(y=1\), the value \(y=1\) must be excluded from both ranges. Explain why this reasoning is incorrect.

Solution 5

The two students produce different parabolas:

\[ \boxed{y=\frac{(x-2)^2}{4}} \]

and:

\[ \boxed{y=\frac{(x-1)^2}{4}} \]

They have different domains, holes, and vertices, but they intersect at exactly one point:

\[ \boxed{\left(\frac32,\frac1{16}\right)} \]

Both transformed graphs have the same range, \([0,9]\).

(a) Final function rules

Start with:

\[ y=x^2 \]

For Student A, the first transformation is a translation one unit right.

Replace \(x\) with \(x-1\):

\[ y=(x-1)^2 \]

The second transformation is a horizontal dilation by a factor of 2.

Replace \(x\) with \(x/2\):

\[ y=\left(\frac{x}{2}-1\right)^2 \]

Simplifying:

\[ \boxed{y=\frac{(x-2)^2}{4}} \]

For Student B, apply the horizontal dilation first:

\[ y=\left(\frac{x}{2}\right)^2 \]

Now translate one unit right by replacing \(x\) with \(x-1\):

\[ y=\left(\frac{x-1}{2}\right)^2 \]

Therefore:

\[ \boxed{y=\frac{(x-1)^2}{4}} \]

The equations are different, even though both students used the same two transformations.

(b) Domains and ranges

First consider the original domain:

\[ [-2,3]\setminus\{1\} \]

The original graph has range:

\[ [0,9] \]

Removing \(x=1\) does not remove \(y=1\) from the range because:

\[ f(-1)=(-1)^2=1 \]

The point \((-1,1)\) is still included.

Now consider Student A.

The original horizontal coordinate \(u\) moves according to:

\[ x=2(u+1) \]

When \(u=-2\):

\[ x=2(-2+1)=-2 \]

When \(u=3\):

\[ x=2(3+1)=8 \]

The excluded input \(u=1\) moves to:

\[ x=2(1+1)=4 \]

Therefore:

\[ \boxed{D_A=[-2,8]\setminus\{4\}} \]

Since the transformation is horizontal only, the range is unchanged:

\[ \boxed{R_A=[0,9]} \]

For Student B, the original horizontal coordinate moves according to:

\[ x=2u+1 \]

When \(u=-2\):

\[ x=-3 \]

When \(u=3\):

\[ x=7 \]

The excluded input \(u=1\) moves to:

\[ x=3 \]

Therefore:

\[ \boxed{D_B=[-3,7]\setminus\{3\}} \]

Again, the range is unchanged:

\[ \boxed{R_B=[0,9]} \]

(c) Positions of the holes

The original hole is:

\[ (1,1) \]

For Student A, the horizontal mapping is:

\[ x=2(u+1) \]

Therefore:

\[ (1,1)\longrightarrow(4,1) \]

The hole is:

\[ \boxed{(4,1)} \]

For Student B:

\[ x=2u+1 \]

Therefore:

\[ (1,1)\longrightarrow(3,1) \]

The hole is:

\[ \boxed{(3,1)} \]

The holes are in different positions because the horizontal transformations were applied in different orders.

(d) Vertices

The original vertex is:

\[ (0,0) \]

For Student A:

\[ x=2(0+1)=2 \]

Therefore, the vertex becomes:

\[ \boxed{(2,0)} \]

For Student B:

\[ x=2(0)+1=1 \]

Therefore, the vertex becomes:

\[ \boxed{(1,0)} \]

Both parabolas open upwards, but their vertices are in different positions.

(e) Intersection points

To find where the graphs intersect, set their equations equal:

\[ \frac{(x-2)^2}{4} = \frac{(x-1)^2}{4} \]

Multiply both sides by 4:

\[ (x-2)^2=(x-1)^2 \]

Expand:

\[ x^2-4x+4=x^2-2x+1 \]

Cancel \(x^2\):

\[ -4x+4=-2x+1 \]

Rearrange:

\[ 3=2x \]

Therefore:

\[ \boxed{x=\frac32} \]

Substitute into either equation:

\[ y=\frac{\left(\frac32-2\right)^2}{4} \]

\[ =\frac{\left(-\frac12\right)^2}{4} \]

\[ =\frac{1/4}{4} \]

Therefore:

\[ \boxed{y=\frac1{16}} \]

The intersection is:

\[ \boxed{\left(\frac32,\frac1{16}\right)} \]

We must check that this point belongs to both domains.

For Student A:

\[ -2\leq\frac32\leq8 \]

and \(\frac32\neq4\).

For Student B:

\[ -3\leq\frac32\leq7 \]

and \(\frac32\neq3\).

Therefore, the intersection is valid for both graphs.

There are no other intersection points because the equation reduces to a linear equation with exactly one solution.

(f) Does the missing point remove \(y=1\) from the range?

No. The value \(y=1\) remains in both ranges.

This is the most subtle part of the question.

The original graph has a hole at:

\[ (1,1) \]

However, it also contains:

\[ (-1,1) \]

Removing the first point does not remove the second.

After Student A’s transformations, the included point \((-1,1)\) moves to:

\[ x=2(-1+1)=0 \]

Therefore, the transformed graph still contains:

\[ (0,1) \]

After Student B’s transformations, the same original point moves to:

\[ x=2(-1)+1=-1 \]

Therefore, Student B’s graph still contains:

\[ (-1,1) \]

The tempting mistake is to assume that removing one point from a graph always removes its corresponding \(y\)-value from the range.

That is only true when no other allowed input produces the same output.

Here, the symmetry of the original parabola means that both \(x=1\) and \(x=-1\) produce \(y=1\). Removing one of them leaves the other.

This is also why tracking the domain and range as sets is more reliable than simply copying the coordinates of a hole.

13Where to go next

Graph transformations become much more useful when you can combine them with other function skills.

The next step is to apply the same coordinate-mapping approach to inverse functions, composite functions, and transformations involving restricted domains.

In particular, understanding how transformations move asymptotes, intercepts, and discontinuities will help when you begin analysing more complicated rational and logarithmic graphs in Mathematics Advanced.

The underlying method stays the same: start with a point, track how its coordinates change, and use those changes to reconstruct the graph.