Lesson Explainer: Function Transformations: Reflection Mathematics

In this explainer, we will learn how to reflect a graph on the π‘₯- or 𝑦-axis, both graphically and algebraically.

One of the fundamental concepts in geometry is that of transforming a shape with the standard actions of translation, rotation, reflection, and dilation. Often, these concepts are illustrated using polygons and other common concepts, usually with familiar and ubiquitous shapes, such as triangles and circles. Once these notions are understood intuitively, it is common to begin treating the subject a little more precisely, with the aim of understanding exactly what happens to a shape when it undergoes some combination of transformations. A common way of illustrating this is to refer to the vertices of a shape, which can be expressed using precise coordinates and hence can have these movements tracked as the result of transformations being applied. Approaching the topic in such a way will allow for a visual understanding of transformations to be combined with concepts that are drawn from coordinate geometry.

In many senses, understanding the effects of transformation on a function can be thought of as a generalization of the above approach. If a function is well-defined (either algebraically or with a suitably descriptive graph), then its qualitative behavior can be known at all points, and we might then be interested in how the function behaves when it is subjected to various transformations. Given that a function is ideally written as a formula or an algebraic expression, it is a natural extension to ask how transforming the function can be represented within this framework. Fortunately, many transformations are simple to explain using intuitive algebraic rules, especially for some types of translations, reflections, and dilations.

This explainer will focus on what happens to a graph when it is reflected in the π‘₯-axis or 𝑦-axis. It is possible to reflect any function through any straight line, such as the line 𝑦=π‘₯ or any other of the form 𝑦=π‘šπ‘₯+𝑏. These can all be understood algebraically as well as visually, although the simplest case is the reflection in either of the axes. As we will see, these two types of reflections are simple to understand visually, and have a similarly straightforward algebraic interpretation.

Definition: Reflecting a Function in the Horizontal or Vertical Axis

Consider a function 𝑦=𝑓(π‘₯) that is plotted on the standard π‘₯𝑦-axis. Then, a reflection in the π‘₯-axis can be produced using the function 𝑦=βˆ’π‘“(π‘₯), and a reflection in the 𝑦-axis can be found using the function 𝑦=𝑓(βˆ’π‘₯).

As an aside, we might ask to verify the above definition with a simple check. For example, we would expect that if we were to reflect a function 𝑓(π‘₯) in the π‘₯-axis and then reflect the new function again in the π‘₯-axis, then we would recover the original function. The argument would similarly hold for a double reflection in the 𝑦-axis. We can see that this property is respected by the algebraic rules that were set out in the definition above. Suppose we took some function 𝑓(π‘₯) and reflected in the π‘₯-axis to obtain 𝑔(π‘₯). Using the definition would give 𝑔(π‘₯)=βˆ’π‘“(π‘₯). If we were then to reflect 𝑔(π‘₯) in the π‘₯-axis to obtain a new function β„Ž(π‘₯), then we would find that β„Ž(π‘₯)=βˆ’π‘”(π‘₯)=βˆ’(βˆ’π‘“(π‘₯))=𝑓(π‘₯), thus returning the original function, as expected. Similar working shows that the same result holds when reflecting a function twice in the 𝑦-axis.

We will demonstrate both of these types of reflections using an example quadratic function. In particular, we will use the function 𝑓(π‘₯)=π‘₯βˆ’4π‘₯+3.

This function has two roots that can be found by factoring or by the quadratic root formula as π‘₯=1 and π‘₯=3. Plotting this function in the range π‘₯∈[βˆ’5,5] produces the graph below, which we could verify by creating a table of values or by using any standard graph-plotting software. Note that we have marked the roots with red dots.

We will now seek to reflect this function in the π‘₯-axis, which will mean reflecting about the horizontal line 𝑦=0. Given that the function is fairly simplistic, we can predict that the effect will be as shown in the graph below. Note that the roots are unchanged and we, in essence, have just β€œflipped” the function about the horizontal axis.

As stated in the definition above, we can understand this transformation in an algebraic sense by creating the function 𝑦=βˆ’π‘“(π‘₯). This amounts to little more than an overall sign change in every term of the function and, given that we defined 𝑓(π‘₯)=π‘₯βˆ’4π‘₯+3, when reflected in the π‘₯-axis, we would simply take βˆ’π‘“(π‘₯)=βˆ’ο€Ήπ‘₯βˆ’4π‘₯+3=βˆ’π‘₯+4π‘₯βˆ’3.

Should we wish to, we could verify that this algebraic expression does indeed produce the graph shown above and that the roots are unchanged. As a way of confirming our result, we can see now that the 𝑦-intercept term has had a change of sign, which is expected from a reflection in the π‘₯-axis. We can also see that the shape of the quadratic function has shifted from a β€œu” shape to an β€œn” shape, as respected by the change in sign on the π‘₯-term.

This type of reflection is normally thought of as being the slightly easier of the two. Geometrically speaking, both types of reflections are easy to visualize, although when we reflect in the 𝑦-axis, we need to be a little more careful. Suppose we took the original function and decided to reflect in the 𝑦-axis, which would mean that we would be reflecting in the vertical line π‘₯=0. Applying this transformation to the original function would return the following graph.

Note that the 𝑦-intercept is unchanged but that the roots have changed sign. The overall shape of the quadratic is still a β€œu” shape, which means that the quadratic term must have a positive coefficient. To understand how to express this new function algebraically and explicitly, we can make the substitution π‘₯β†’βˆ’π‘₯ in the original function. Doing this results in the following working: 𝑓(βˆ’π‘₯)=(βˆ’π‘₯)βˆ’4(βˆ’π‘₯)+3=π‘₯+4π‘₯+3.

This result is of the form that we expected, with the 𝑦-intercept being unchanged along with the sign of the quadratic term. The only term that has changed is that of the linear term, although this is enough to cause the change in shape that we have witnessed. We can check that the plot corresponds to the equation above and that the roots are exactly π‘₯=βˆ’1 and π‘₯=βˆ’3.

Although we have chosen to use a quadratic function to demonstrate these two reflections, we easily could have chosen any other type of function that could be described explicitly. Often, such concepts are demonstrated using polynomials (typically linear, quadratic, or cubic), although we can use any function for this purpose. We will now begin to cover a series of examples that will illustrate the ideas we have discussed above.

Example 1: Identifying the Graph of a Function Following a Reflection in an Axis

Consider the function 𝑓(π‘₯)=(π‘₯βˆ’1)+2, where π‘₯=[1,∞[.

  1. Which of the graphs represents the function 𝑓(π‘₯)?
  2. Which of the graphs shown represents the reflection of 𝑓(π‘₯) on the π‘₯-axis?
  3. Which of the graphs shown represents the reflection of 𝑓(π‘₯) on the 𝑦-axis?


Part 1

We are told that π‘₯∈[1,∞[, which means that the function is not defined when π‘₯ takes values less than 1. This means that we can exclude the options B and D, which both appear to be defined in the region π‘₯∈]βˆ’βˆž,βˆ’1].

Next, we can try investigating the behavior of the function 𝑓(π‘₯) at a particular value. It is normally sensible to begin with boundary values, and in this case, we will consider the situation when π‘₯=1. Our definition of 𝑓(π‘₯) gives 𝑓(1)=(1βˆ’1)+2=2, which means that the graph of the function must pass through the point (1,2). From the remaining options of A, C, and E, the only graph that has this property is option C.

Part 2

We have already done most of the hard work by determining that 𝑓(π‘₯) corresponds to option C. By checking the graph above, we can see that reflecting in the π‘₯-axis gives the graph A and that there are no other possibilities. Given that the new function would be determined by 𝑦=βˆ’π‘“(π‘₯) and since 𝑓(π‘₯) is entirely positive in the given domain, the reflected function must be entirely negative in the same domain. This confirms that the only option is A.

Part 3

Looking at the original function 𝑓(π‘₯) plotted by option C, reflecting in the 𝑦-axis clearly gives option B. To confirm this, recall that the reflected function would be found by plotting 𝑦=𝑓(βˆ’π‘₯). We could confirm this by recalling that 𝑓(π‘₯) is defined with the domain π‘₯∈[1,∞[, which must mean that the reflected function is defined under the domain π‘₯∈]βˆ’βˆž,βˆ’1]. The only two options with this domain are B and D. Given that 𝑓(π‘₯) is entirely positive, this must also be the case for 𝑦=𝑓(βˆ’π‘₯), which eliminates D as an option.

It is expected that students will be able to perform a reflection in the π‘₯-axis or 𝑦-axis by recognizing the two transformations 𝑦=βˆ’π‘“(π‘₯) and 𝑦=𝑓(βˆ’π‘₯) without it necessarily being mentioned that this corresponds to a particular reflection. Sometimes, you will be asked to visually depict these transformations without actually being given an explicit formula for the function itself or without being told directly that you are effectively being asked to perform a reflection. We will see this in the next example.

Example 2: Identifying the Graph of a Function Following a Reflection Which Is Given Algebraically

This is the graph of 𝑦=𝑔(π‘₯).

Which of the following is the graph of 𝑔(βˆ’π‘₯)?


We know that the function 𝑔(βˆ’π‘₯) will correspond to a reflection in the 𝑦-axis. In other words, we will need to reflect about the line π‘₯=0. This will incur a sign change in the roots but will leave the 𝑦-intercept unchanged. In 𝑔(π‘₯), there appears to be one root at π‘₯=βˆ’3 and a double root at π‘₯=2, meaning that the function 𝑔(βˆ’π‘₯) should have a single root at π‘₯=3 and a double root at π‘₯=βˆ’2. The only option obeying this property as well as maintaining the 𝑦-intercept is option (b).

To expound upon the previous example, it is worth noting that the graph in option (a) has clearly been obtained by reflecting the function 𝑔(π‘₯) in the π‘₯-axis, which corresponds to the graph 𝑦=βˆ’π‘”(π‘₯). As an exercise left to the reader, it appears that the graph of option (c) is the original function 𝑔(π‘₯) after having been reflected in the π‘₯-axis and then again in the 𝑦-axis (or vice versa). This function would be plotted using the transformation 𝑦=βˆ’π‘”(βˆ’π‘₯), and we will discuss this matter again in a later example.

Example 3: Finding the Equation of a Function When Given the Graph of Its Reflection in an Axis

The following linear graph represents a function 𝑔(π‘₯) after a reflection in the 𝑦-axis. Find the original function 𝑓(π‘₯).


The solution of this problem is divided into two stages. First, we determine the equation of the line 𝑔(π‘₯) that is plotted in the graph above. Then, we use known algebraic properties to recover the function 𝑓(π‘₯). We are told that 𝑔(π‘₯) has been obtained by reflecting the function 𝑓(π‘₯) in the 𝑦-axis. This means that we have reflected about the line π‘₯=0, hence, giving the relation 𝑓(π‘₯)=𝑔(βˆ’π‘₯) that will be useful later.

The function 𝑔(π‘₯) clearly has a 𝑦-intercept of βˆ’4 and a single root when π‘₯=βˆ’2. Phrased in terms of coordinates, this means that the straight line passes through the points (0,βˆ’4) and (βˆ’2,0), which we will label as (π‘₯,𝑦) and (π‘₯,𝑦) respectively. There are multiple ways of calculating the equation of this straight line given the information at hand, but one method is to solve the following equation for 𝑦: π‘¦βˆ’π‘¦π‘¦βˆ’π‘¦=π‘₯βˆ’π‘₯π‘₯βˆ’π‘₯.

Substituting in the known values gives π‘¦βˆ’(βˆ’4)0βˆ’(βˆ’4)=π‘₯βˆ’0βˆ’2βˆ’0, which can be solved to give 𝑦=βˆ’2π‘₯βˆ’4. This means that the function of interest is 𝑔(π‘₯)=βˆ’2π‘₯βˆ’4. This completes the first part of the question.

We can now complete the second part of the question by recalling that the function 𝑔(π‘₯) was derived from another function 𝑓(π‘₯) after a reflection in the 𝑦-axis. In other words, we have 𝑓(π‘₯)=𝑔(βˆ’π‘₯). By substituting π‘₯β†’βˆ’π‘₯ into 𝑔(π‘₯), we obtain 𝑓(π‘₯)=𝑔(βˆ’π‘₯)=βˆ’2(βˆ’π‘₯)βˆ’4=2π‘₯βˆ’4.

Note that the value of the 𝑦-intercept has not changed but that there has been a sign change in the single root. The only root of 𝑓(π‘₯) is when π‘₯=2, whereas the only root of 𝑔(π‘₯) is π‘₯=βˆ’2.

Problems such as in the previous example are helpful for consolidating multiple areas of mathematics into a single overarching process. In the previous example, we had to find the equation of a straight line from two points, then use the correct algebraic substitution to obtain a suitable reflection, and then produce a new function as a result of this. Provided we are working with straight lines, this would never be an overly tricky experience, as many methods of finding the equation of the straight line are available. However, performing the actual algebraic substitution and then deriving the new function is seldom much more difficult than in the previous example. In the following example, we again look to solve such a problem but this time with a quadratic function whose equation must be found before the remainder of the question can be answered. Once again, there are many possible techniques for obtaining the exact expression of the quadratic function, which are largely dependent on preference.

Example 4: Finding the Equation of a Function When Given the Graph of Its Reflection in an Axis

The following parabolic graph represents a function 𝑔(π‘₯) after a reflection on the π‘₯-axis. Find the original function 𝑓(π‘₯).


We will answer this question in two parts. First, we will find the quadratic function 𝑔(π‘₯) as plotted in the graph above. Then, we will use our knowledge of function transformations to find an expression for 𝑓(π‘₯).

There are multiple ways that we could approach finding the equation of the quadratic function in the above plot. We can see that the function has a 𝑦-intercept of βˆ’6, meaning that we can use the coordinate (0,βˆ’6). Additionally, it appears as though there is a local maximum at the coordinate (1,βˆ’5). We can also infer that the curve passes through the point (2,βˆ’6). We can use these three coordinates to recover the equation of 𝑔(π‘₯).

Given that 𝑔(π‘₯) is a parabola (another name for a quadratic), we will begin by recalling the general form of a quadratic: 𝑔(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+𝑐, where π‘Ž, 𝑏, and 𝑐 are real numbers. We know that the graph passes through the point (0,βˆ’6), meaning that we must have 𝑔(0)=βˆ’6. Substituting into the general quadratic equation above gives 𝑔(0)=π‘ŽΓ—0+𝑏×0+𝑐=βˆ’6, giving the result 𝑐=βˆ’6.

With this knowledge in hand, we can now use the next known coordinate of (1,βˆ’5). Accordingly, we must have 𝑔(1)=βˆ’5, which gives 𝑔(1)=π‘ŽΓ—1+𝑏×1βˆ’6=βˆ’5, where we have substituted the known result 𝑐=βˆ’6. Simplifying gives π‘Ž+𝑏=1.

This equation features the unknown coefficients π‘Ž and 𝑏, and we can actually obtain a second equation featuring these coefficients using the known coordinate (2,βˆ’6). We set 𝑔(2)=βˆ’6 to give 𝑔(2)=π‘ŽΓ—2+𝑏×2βˆ’6=βˆ’6.

Rearranging gives 4π‘Ž+2𝑏=0. We can then use this equation along with the equation π‘Ž+𝑏=1 to solve these simultaneous equations and hence find that π‘Ž=βˆ’1 and 𝑏=2.

Now that π‘Ž, 𝑏, and 𝑐 are known, we have 𝑔(π‘₯)=βˆ’π‘₯+2π‘₯βˆ’6. By completing the square, or by using knowledge of how the π‘₯-value of the minimum/maximum of a quadratic is related to the coefficients π‘Ž, 𝑏, and 𝑐, we can hence derive the more informative expression 𝑔(π‘₯)=βˆ’(π‘₯βˆ’1)βˆ’5, which actually confirms that the maximum occurs when π‘₯=1. Now that we have found 𝑔(π‘₯), we can obtain 𝑓(π‘₯) given that this function is found after reflecting 𝑔(π‘₯) in the π‘₯-axis. To find the exact formula for 𝑓(π‘₯), we use the corresponding relation: 𝑓(π‘₯)=βˆ’π‘”(π‘₯). This gives the final answer: 𝑓(π‘₯)=(π‘₯βˆ’1)+5.

Earlier, we alluded to the idea of reflecting a function in the π‘₯-axis and then the 𝑦-axis (or vice versa). Supposing we had a function 𝑓(π‘₯), we know that we could obtain a new function 𝑔(π‘₯) by reflecting 𝑓(π‘₯) in the π‘₯-axis by setting 𝑔(π‘₯)=βˆ’π‘“(π‘₯). If we were to take this new function 𝑔(π‘₯) and then choose to reflect it in the 𝑦-axis to obtain another new function β„Ž(π‘₯), then we would be setting π‘₯β†’βˆ’π‘₯ into 𝑔(π‘₯). In other words, we would have β„Ž(π‘₯)=𝑔(βˆ’π‘₯)=βˆ’π‘“(βˆ’π‘₯), which we can simply write in brief as β„Ž(π‘₯)=βˆ’π‘“(βˆ’π‘₯). Had we performed the steps in the reverse order (by first reflecting in the 𝑦-axis and then reflecting in the π‘₯-axis), we would have obtained exactly the same result, albeit with slightly different interim steps.

Example 5: Understanding Combined Reflections of Functions

Which of the following processes would you use to obtain the graph of 𝑦=βˆ’π‘“(βˆ’π‘₯) from the graph 𝑦=𝑓(π‘₯)?

  1. Reflecting the graph in the π‘₯-axis
  2. Reflecting the graph in the 𝑦-axis
  3. Reflecting the graph in the line 𝑦=π‘₯
  4. Reflecting the graph in the line 𝑦=βˆ’π‘₯


If we consider the function 𝑦=βˆ’π‘“(βˆ’π‘₯), we can break it down into two separate transformations of 𝑦=𝑓(π‘₯). The first is the transformation 𝑔(π‘₯)=𝑓(βˆ’π‘₯). Here, 𝑔(π‘₯) is a reflection of 𝑓(π‘₯) in the 𝑦-axis. The second is β„Ž(π‘₯)=βˆ’π‘”(π‘₯), which represents a reflection of 𝑔(π‘₯) in the π‘₯-axis. Notice also that β„Ž(π‘₯)=βˆ’π‘“(βˆ’π‘₯). Therefore, this new graph is achieved by first reflecting in the 𝑦-axis and then reflecting in the π‘₯-axis (or vice versa).

The previous example introduced the idea of reflecting in lines that are neither the π‘₯-axis or the 𝑦-axis. The main way of approaching such problems would involve expressing the original function and then using a matrix to transform the function as needed. This is clearly beyond the scope of the explainer, but it is worth mentioning that reflecting in different lines will have a highly different algebraic structure that is not always possible to express. For example, reflecting a graph in the line 𝑦=π‘₯ amounts to finding the inverse of the original function, which is not always viable. In fact, it is only generally possible if the original function is a one-to-one function.

Example 6: Understanding the Algebraic Representation of a Reflection in an Axis

Find 𝑔(π‘₯), where the graph of the function 𝑔(π‘₯) is a reflection across the π‘₯-axis of the graph of the function 𝑓(π‘₯)=2π‘₯+5.


Note that we do not need to plot this graph in order to find how to express it algebraically. We know that 𝑓(π‘₯)=2π‘₯+5 and that 𝑔(π‘₯) is the same function after a reflection in the π‘₯-axis. This means that 𝑔(π‘₯)=βˆ’π‘“(π‘₯) and hence we find the answer: 𝑔(π‘₯)=βˆ’π‘“(π‘₯)=βˆ’(2π‘₯+5)=βˆ’2π‘₯βˆ’5.

So far in this explainer, we have worked with functions that are expressed by means of a specific algebraic formula. We have seen generally that for a function 𝑓(π‘₯), a reflection in the π‘₯-axis can be found by creating the new function 𝑔(π‘₯)=βˆ’π‘“(π‘₯) and also that a reflection in the 𝑦-axis can be found by creating the new function β„Ž(π‘₯)=𝑓(βˆ’π‘₯). Given that these relationships hold true for all values of π‘₯, it must be the case that they also hold true for specific values. In other words, if we are given a table of values from a particular function, then we should be able to use the known results to understand how to reflect this function in either the π‘₯-axis or the 𝑦-axis.

For example, suppose we have the function 𝑓(π‘₯), which is defined by the following table of values.


We can plot these coordinates as points by setting 𝑦=𝑓(π‘₯), without attempting to join the points with any form of curve or straight line, as shown in the figure below. Even though this function is not continuous, we can still reflect it in the π‘₯-axis or the 𝑦-axis.

Recall that reflecting in the π‘₯-axis is the same as reflecting in the line 𝑦=0, meaning that any roots of the function will not be affected. We create a new function that represents this reflection and we label it 𝑔(π‘₯). We know that 𝑔(π‘₯)=βˆ’π‘“(π‘₯), which must be true for every known value. In other words, we know that 𝑔(βˆ’2)=βˆ’π‘“(βˆ’2), 𝑔(βˆ’1)=βˆ’π‘“(βˆ’1), 𝑔(0)=βˆ’π‘”(0), 𝑔(1)=βˆ’π‘”(1), and 𝑔(2)=βˆ’π‘”(2). This allows us to create the new table of values.


Plotting these points gives the following graph, which we can see is the same as the original graph after a reflection in the π‘₯-axis. Note that the point (1,0) is unaffected because it sits on the line π‘₯=0, which we have just reflected through. This is analogous to saying that the roots of a continuous function are unaffected by reflection in the π‘₯-axis.

We will now demonstrate how to complete a reflection in the 𝑦-axis. We create a new function β„Ž(π‘₯) and then define β„Ž(π‘₯)=𝑓(βˆ’π‘₯), which will give the reflection that we are seeking. This means that β„Ž(βˆ’2)=𝑓(2), β„Ž(βˆ’1)=β„Ž(1), β„Ž(0)=β„Ž(0), β„Ž(1)=𝑓(βˆ’1), and β„Ž(2)=𝑓(βˆ’2). Using the initial table of values for 𝑓(π‘₯), we can then produce the new table of values for β„Ž(π‘₯) as follows.


By plotting these points, we obtain the graph below, which is indeed the original function 𝑓(π‘₯) after being reflected in the 𝑦-axis. The point (0,3) is unchanged by a reflection in the 𝑦-axis because this is a reflection in the line π‘₯=0 on which this point sits. This is analogous to saying that the 𝑦-intercept of a continuous function is unchanged after a reflection in the 𝑦-axis.

Example 7: Using a Table of Values to Identify the Reflection of a Function in an Axis

Consider the following table of the function 𝑓(π‘₯).


Choose the table of the reflected function 𝑔(π‘₯) over the 𝑦-axis.

  1. π‘₯1234
  2. π‘₯βˆ’1βˆ’2βˆ’3βˆ’4
  3. π‘₯1234
  4. π‘₯βˆ’1βˆ’2βˆ’3βˆ’4
  5. π‘₯1βˆ’234


We will begin by plotting the values in the table as points in the coordinate plane, as shown on the left-hand side in the diagram. The function after reflection in the 𝑦-axis, which is labeled 𝑔(π‘₯), is plotted on the right-hand side.

This information alone should be enough to answer the question fully, but we will also demonstrate the process more precisely. We recall that reflecting 𝑓(π‘₯) in the 𝑦-axis to obtain a new function 𝑔(π‘₯) is understood algebraically by the relationship 𝑔(π‘₯)=𝑓(βˆ’π‘₯). This must be true for all values of π‘₯, and we only have outputs for the four input values π‘₯=1, 2, 3, and 4. This means that the only output values that we can obtain for 𝑔(π‘₯) are for the four input values π‘₯=βˆ’1, βˆ’2, βˆ’3, and βˆ’4. We therefore find that 𝑔(βˆ’1)=𝑓(1), 𝑔(βˆ’2)=𝑓(2), 𝑔(βˆ’3)=𝑓(3), and 𝑔(βˆ’4)=𝑓(4). Entering this into the new table of values for 𝑔(π‘₯), we have the following.


This corresponds to option D.

Key Points

  • Consider a function 𝑦=𝑓(π‘₯). A reflection in the π‘₯-axis has the algebraic expression 𝑦=βˆ’π‘“(π‘₯), and a reflection in the 𝑦-axis is represented as 𝑦=𝑓(βˆ’π‘₯).
  • Reflecting twice in either the π‘₯-axis or the 𝑦-axis will return the original function.
  • Reflecting in the π‘₯-axis does not change the roots of a function. However, this will change the sign of the 𝑦-intercept.
  • Reflecting in the 𝑦-axis will change the signs of the roots, but it will not change the sign of the 𝑦-intercept.
  • A combined reflection in the π‘₯-axis and then the 𝑦-axis (or vice versa) is represented as 𝑦=βˆ’π‘“(βˆ’π‘₯).
  • When reflecting in lines other than the π‘₯-axis or 𝑦-axis, other approaches are needed and the algebraic interpretation is not so straightforward.

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