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Graphical Transformations in Year 11 Maths Extension 1 builds on both prior basic polynomial sketching and sketching basic graphs in junior years.
In senior courses, being able to confidently sketch and transform a graph(s) is a key skill. These include shifting the graph, reflecting the graph, graphs involving squares and square roots and also addition, subtraction and multiplication of two functions. These will be essential skills that are necessary for applications later in Year 11 and 12, where sketching a curve enables for a visual representation of complex function problems.
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NESA requires students to be proficient in the following outcomes:
Students should have a basic understanding of curve sketching and functions. This includes having a solid grasp on the concepts of \(x\)-intercepts and \(y\)-intercepts as well as vague knowledge on the various transformations which can be applied to graphs.
This content can be found in our Year 11 Advanced Guides should students want to solidify their understanding:
In this topic, we aim to obtain rough sketches of graphs with minimal use of calculus. In many curve sketching questions, we will not be provided with an equation. Instead we would be provided with a diagram and a generic transformation to apply. Below are the different types of basic transformations that could be asked:
For questions which involve a mix of these transformations, always decompose the question into the basic transformations and apply in the order: horizontal changes (transformation then scaling), reflections, absolute value transformations and lastly vertical changes (transformation then scaling).
For all of our examples, we will be referencing this graph:
The curve given by the equation \(y = f(x)+c\), is just the curve \(y=f(x)\) shifted vertically by \(c\) units.
Example 1
On the axis below, draw the graph of \(y=f(x)-2\)
Solution 1
For this transformation, as \(c<0\), we shift the curve down \(2\) units:
The curve given by the equation \(y = f(x-c)\), is just the curve \(y=f(x)\) shifted horizontally by \(c\) units.
Example 2
On the axis below, draw the graph of \(y=f(x+\frac{1}{2}\)
Solution 2
For this transformation, as \(c<0\), we shift the curve right \(c\) units:
The curve given by the equation \(y = cf(x)\), is just the curve \(y=f(x)\) scaled vertically by \(c\) units.
Importantly, if the question only involves vertical scaling, only the y-intercepts are scaled, not the \(x\)-intercepts.
Example 3
On the axis below, draw the graph of \(y=2f(x)\)
Solution 3
For this transformation, as \(c>1\), we stretch vertically by a factor of \(2\):
The curve given by the equation \(y = f(cx)\), is just the curve \(y=f(x)\) scaled horizontally by \(c\) units.
With horizontal scaling, it is a little bit more difficult as we are often going against our intuition. Importantly, if the question only involves horizontal scaling, only the \(x\)-intercepts are scaled, not the \(y\)-intercepts.
Example 4
On the axis below, draw the graph of \(y=f(3x)\)
Solution 4
For this transformation, as \(c>1\), we squash horizontally by a factor of \(3\):
Questions which require a reflection about the axes are easily denoted by the inclusion of a negative sign:
Example 5
On the axis below, draw the graph of \(y=f(-x)\).
Solution 5
As we have the negative sign right next to the \(x\), this is a reflection about the \(y\)-axis:
There are only 2 types of absolute value transformations which can be asked in Year 11 Extension 1 Mathematics:
If there is a question which requires both absolute value transformations, we always apply them in the order listed above.
Example 6
On the axis below, draw the graph of \(y=|f(|x|)|\).
Solution 6
First, we apply the \(y=|f(x)|\) transformation. Therefore, we must reflect any part of the graph which is below the \(x\)-axis:
Then, we apply the \(y=f(|x|)\). Therefore, we remove any part of the graph which is left of the \(y\)-axis and reflect the part which is to right of the \(y\)-axis. This leaves us with the complete \(y=|f(|x|)|\) transformation:
Harder transformations of graphs involve graphing square roots and the addition, subtraction and multiplication of two functions. Graphing such functions can be approached by splitting the graph into segments containing key features, then combining to form the transformed graph.
When sketching the square root of a given function, we should consider key features within the original function and how they would appear once we square root the function. Here is a reference table of the transformations to a square root function.
| Features of \(y=f(x)\) | Features of \(y=\sqrt{f(x)}\) |
| \(f(x)<0\) | Square rooting negative values results in non-real values. Therefore there are no values. |
| \(0<f(x)<1\) | Square rooting values within this range yields values that are larger. E.g. \(\sqrt{\frac{1}{4}}=\frac{1}{2}\) |
| \(f(x)=1\) | Remains the same value. |
| \(f(x)>1\) | Results in smaller values |
| Vertical Asymptotes | Unchanged. |
| Horizontal Asymptotes | Produced a new horizontal asymptote at the square rooted value. |
| Increasing values of \(f(x)\) | Remains increasing |
| Decreasing values of \(f(x)\) | Remains decreasing |
By splitting the graph into the ranges, we are able to sketch each segment with reference to how the features are transformed. By splitting the graph into the ranges, we are able to sketch each segment with reference to how the features are transformed.
Example 1
In this example, we are given \(y=f(x)\) where \(g(x)=\sqrt{f(x)}\). To approach any graph, we can split the range into 4 segments.
We then have to consider some key features of the graph. All increasing, decreasing portions as well as stationary points above the \(x\)-axis are retained. Asymptotes also have to be considered. From the reference table, horizontal asymptotes are square rooted. However, as the value of the asymptote is \(y=1\), this would not change.
Graphing \(y^2=f(x)\) is very similar to graphing the square root graph. We can square root both sides of the equation to yield the following same result:
From the previous section on reflections, we have learnt that \(y=-f(x)\) would reflect the graph along the \(x\)-axis. As the equation has both the positive and negative function, we would have the graph both.
Example 2
From the example in Graphs of \(y=\sqrt{f(x)}\) to graph \(y^2=f(x)\) we would just simply add the reflection of the square root graph across the \(x\)-axis.
Note: Remember that asymptotes are also reflected and need to be shown in the graph.
To graph reciprocal graphs, we must first consider the key features and how these are transformed.
| Features of \(y=f(x)\) | Features of \(y=\frac{1}{f(x)}\) |
| \(x\)-intercept | |
| Increasing values of \(f(x)\) | Becomes decreasing values of \(f(x)\) |
| Decreasing values of \(f(x)\) | Becomes increasing values of \(f(x)\) |
| \(f(x)=±1\) | Remains the same |
| Large values of \(f(x)\) | Become small |
| Small values of \(f(x)\) | Become large |
| As \(x→∞\), \(y→∞\) | As \(x→∞\), \(y→0\) |
Example 3
To transform the graph to its reciprocal graph, we can determine the key features and transform them before beginning to sketch.
Solution 3
\(y=f(x)\) has an \(x\)-intercept at \(x=0\). As the reciprocal produces an undefined value, this becomes an asymptote on our reciprocal graph. This has the equation \(x=0\). The horizontal asymptote should be transformed such that it is the reciprocal value. Since the equation is \(x=1\), it remains the same.
We then need to consider the limits as \(x\) approaches \(±∞\) and to the left and right of vertical asymptotes.
Taking into account that increasing values of \(f(x)\) become decreasing values of \(f(x)\) and vice versa, combined with the key features we have drawn before, we are able to construct the reciprocal graph.
To graph the addition or subtraction of two functions, we firstly have to graph \(y=f(x)\) and\( y=g(x)\) separately on the same set of axes, adding coordinates. Key features that will aid us include:
| Points of Interest | Features of \(y=f(x)+g(x)\) |
| \(g(a)=0\) | When \(g(a)=0\), then the coordinate point on our addition graph would be \(y=f(a)\) at \(x=a\) |
| \(f(a)=0\) | Similarly, the coordinate point would be \(y=g(a)\) at \(x=a\) |
| \(f(a)=g(a)\) | The value at this point would be twice that of the separate graphs. |
| \(f(a)\) has a vertical asymptote at \(x=a\) and \(g(a)\) exists. | Asymptote remains, however nature of the graph may change. |
Example 4
If we want to graph \(y=f(x)+g(x)\), where \(f(x)=x(x-2)\) and \(g(x)=\frac{1}{x}\), we first have to consider the key features.
Solution 4
\(g(x)\) has a vertical asymptote at \(x=0\). Therefore, we know in our addition of functions, this vertical asymptote would be translated across.
From here, we consider the \(x\)-intercepts of both graphs. On \(f(x)\), there is \(x\)-intercepts at both \(x=0\) and \(x=2\). From the reference table, if \(f(2)=0\), then \(y=g(2)\) as shown by the overlap of the green dotted line (\(y=g(x)\))and red line (\(y=f(x)+g(x)\)) in the diagram below.
We then need to consider the limits as \(x\) approaches \(±∞\) and to the left and right of vertical asymptotes.
If extra points are necessary, simply choose an \(x\) value and add the \(y\) values to achieve the addition graph.
Graphing \(y=f(x)-g(x)\) can be done in two ways.
The key features in a \(y=f(x)g(x)\) graph are outlined below
| Points of Interest | Features of \(y=f(x)g(x)\) |
| \(f(a)=0\) and \(g(a)\) exists | \(x\)-intercept at \(x=a\) |
| \(f(x)\)and \(g(x)\) intersect at \(x=a\) | The \(y\) value would be the square of \(f(x)\) |
| \(f(x)=1\) and \(g(a\)) exists | The \(y\) value would be \(g(a)\) |
When graphing, it is important to consider the sign of the \(y\) values. For example.
\begin{align*}
&\text{Positive} \ × \text{Positive} = \ \text{Positive} \\
&\text{Positive} \ × \text{Negative} = \ \text{Negative} \\
&\text{Negative} \ × \text{Negative} = \ \text{Negative} \\
\end{align*}
Example 5
Graph \(y=f(x)g(x)\) where \(f(x)=\frac{1}{2} x^3\) and \(g(x)=2x\).
Solution 5
Similar to all harder transformation questions, we first have to graph each function individually to identify the key features. Firstly, we can identify the \(x\) intercepts of each function, as this would appear as an \(x\)-intercept on our transformed graph.
Then, we have to consider the limits as \(x\) approaches \(±∞\). From the separate graphs:
\(x→∞\),\(y→∞\)
\(x→-∞\),\(y→-∞\)
By considering the sign of each \(y\) limit, we are able to find the limit of our multiplication graph. As \(x\) approaches positive infinity, both functions approach a positive value of infinity. Therefore, this results in an infinity limit on our multiplication graph. However, as \(x\) approaches negative infinity, both functions approach negative value of infinity.
Multiplication between two negative limits would result in a positive limit as seen on the graph below.
1. For the curve \(y=f(x)\) below, apply the following transformations:
2. For the curve \(y=g(x)\) below, apply the following transformations:
3. Graph \(y=f(x)-g(x)\), where \(f(x)=e^(-x)\) and \(g(x)=e^x\)
4. Graph \(y=f(x)g(x)\), where \(f(x)=e^x\) and \(g(x)=\frac{1}{x}\)
1.
(1)
(2)
2
(1)
(2)
3.
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