Lesson Video: Function Transformations: Reflection Mathematics

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

13:44

Video Transcript

In this video, we will learn how to reflect a graph on the π‘₯- or 𝑦-axis, both graphically and algebraically. It is possible to reflect any function through any straight line of the form 𝑦 equals π‘šπ‘₯ plus 𝑏. However, in this video, we will focus on what happens to a graph when it is reflected in the π‘₯-axis or 𝑦-axis. Let’s begin by considering how they can be interpreted algebraically. If we consider a function 𝑦 which is equal to 𝑓 of π‘₯ that is plotted on the standard π‘₯𝑦-plane, then a reflection in the π‘₯-axis can be produced using the function 𝑦 is equal to negative 𝑓 of π‘₯ and a reflection in the 𝑦-axis can be produced using the function 𝑦 equals 𝑓 of negative π‘₯.

We will now demonstrate both types of reflections using a quadratic function. The quadratic function π‘₯ squared minus four π‘₯ plus three is shown in the graph. Setting the function equal to zero, we have two roots: when π‘₯ equals one and π‘₯ equals three. Reflecting this function in the π‘₯-axis will mean reflecting about the horizontal line 𝑦 equals zero. In practice, we simply flip the graph about the horizontal axis. As already stated, we know that this represents the function negative 𝑓 of π‘₯ as multiplying every value of the function by negative one changes the positive 𝑦-values to negative and vice versa. The points are the same distance from the π‘₯-axis but on the opposite side. In this case, this is equal to the negative of π‘₯ squared minus four π‘₯ plus three.

Simplifying the right-hand side, we have negative π‘₯ squared plus four π‘₯ minus three. We can see from the graph that the roots are unchanged. The roots of negative 𝑓 of π‘₯ are π‘₯ equals one and π‘₯ equals three. We notice that the sign of the quadratic term has gone from positive to negative. We know that this is correct as the original function was U-shaped but the reflected function is an n-shaped graph. We can also reflect the original function in the 𝑦-axis or the line π‘₯ equals zero. This transformation affects the roots. In fact, they change sign. The roots of our new quadratic are π‘₯ equals negative one and π‘₯ equals negative three. We know that this new function is 𝑓 of negative π‘₯ as multiplying every π‘₯-value by negative one changes the positive values to negative and vice versa. The points are the same distance from the 𝑦-axis but on the opposite side.

Replacing the π‘₯-terms in our original function with negative π‘₯, we have negative π‘₯ squared minus four multiplied by negative π‘₯ plus three. This simplifies to π‘₯ squared plus four π‘₯ plus three. This is a U-shaped quadratic graph with roots at negative one and negative three. We will now look at a specific question involving a quadratic function.

Consider the function 𝑓 of π‘₯ is equal to π‘₯ minus one all squared plus two, where π‘₯ is greater than or equal to one and less than ∞. There are three parts to this question. Which of the graphs represents the function 𝑓 of π‘₯? Which of the graphs represents the reflection of 𝑓 of π‘₯ on the π‘₯-axis? And which of the graphs represents the reflection of 𝑓 of π‘₯ on the 𝑦-axis?

On the figure given, there are five graphs labeled (A) to (E). We firstly need to identify which graph represents the function π‘₯ minus one all squared plus two where π‘₯ lies on the left-closed, right-open interval from one to ∞. We can immediately exclude options (B) and (D) as these are defined on the interval from negative ∞ to negative one. To work out which of the other three graphs represents the function, we can investigate the behavior of the function at a particular value. It is sensible to begin with the boundary values. And in this case, we will calculate 𝑓 of one. When π‘₯ is equal to one, 𝑓 of π‘₯ is equal to one minus one all squared plus two. This is equal to two, which means that the graph of the function must pass through the point one, two. This is only true of graph C. The graph which represents the function 𝑓 of π‘₯ is (C).

The second part of our question asks us which of the graph represents the reflection of 𝑓 of π‘₯ in the π‘₯-axis. Reflecting the point one, two in the π‘₯-axis gives us the point one, negative two. It is therefore clear that graph A represents the reflection of 𝑓 of π‘₯ on the π‘₯-axis. Reflecting the point one, two in the 𝑦-axis gives us the point negative one, two. This confirms that graph B is a reflection of 𝑓 of π‘₯ on the 𝑦-axis. Our three answers are (C), (A), and (B).

In our next question, we need to find the correct reflection without an explicit formula for the function.

This is the graph of 𝑦 equals 𝑔 of π‘₯. Which of the following is 𝑔 of negative π‘₯? Is it graph A, graph B, or graph C?

We begin by recalling that the graph of 𝑔 of negative π‘₯ will be a reflection in the 𝑦-axis of the graph of 𝑔 of π‘₯. We can see from the graph of 𝑔 of π‘₯ that this function has roots at π‘₯ equals negative three and π‘₯ equals two. We know that when we reflect a function in the 𝑦-axis, the roots change sign. This means that the roots of 𝑔 of negative π‘₯ will be π‘₯ equals negative two and π‘₯ equals three. We can therefore eliminate option (A). And in fact, this is a reflection of 𝑔 of π‘₯ in the π‘₯-axis. The function 𝑔 of π‘₯ has a 𝑦-intercept approximately equal to 1.2. When reflecting any function in the 𝑦-axis, the 𝑦-intercept remains unchanged. This means that we can also eliminate option (C).

The correct answer is therefore option (B). This is the graph of the function 𝑔 of negative π‘₯ as the 𝑦-intercept has remained unchanged and the roots have changed sign.

In our next question, we need to find the equation of a function given the graph of its reflection in an axis.

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

There are a few ways to approach this problem. One way would be to find the equation of the function 𝑔 of π‘₯ given on the graph. As it is a linear function, we know it can be written in the form 𝑦 equals π‘šπ‘₯ plus 𝑏, where π‘š is the slope and 𝑏 is the 𝑦-intercept. It is clear from the graph that the 𝑦-intercept is negative, four. We know that the slope or gradient of any line is equal to the rise over the run. Choosing the two points negative four, four and negative two, zero which lie on the line, the slope is equal to negative four over two. This is equal to negative two. The function 𝑔 of π‘₯ has equation negative two π‘₯ minus four. We are told that 𝑔 of π‘₯ is a reflection of 𝑓 of π‘₯ in the 𝑦-axis. Therefore, 𝑓 of π‘₯ is also a reflection of 𝑔 of π‘₯ in the 𝑦-axis.

This means that 𝑓 of π‘₯ is equal to 𝑔 of negative π‘₯. The function 𝑓 of π‘₯ is therefore equal to negative two multiplied by negative π‘₯ minus four. This simplifies to two π‘₯ minus four. The original linear function 𝑓 of π‘₯ has equation two π‘₯ minus four. An alternative method would have been to have sketched the reflection of 𝑔 of π‘₯ on the graph. When reflecting a graph in the 𝑦-axis, we know that the 𝑦-intercept remains the same and the roots change signs. This means that 𝑓 of π‘₯ will still intercept the 𝑦-axis at negative four and intercept the π‘₯-axis at positive two. As the slope or gradient of this line is two and the 𝑦-intercept is negative four, the equation of the line is two π‘₯ minus four. This confirms the answer we found using our first method.

We will now consider one final example where we use a table of values to identify the reflection of a function in an axis.

Consider the following table of the function 𝑓 of π‘₯. We have π‘₯-values from one to four and corresponding 𝑓 of π‘₯ values also from one to four. Choose the table of the reflected function 𝑔 of π‘₯ over the 𝑦-axis.

We are given five options, (A) to (E). After clearing some space, we will plot the values in the table as points in the coordinate plane. The function 𝑓 of π‘₯ has four coordinates in the table: the points one, one; two, two; three, three; and four, four. We are told that the function 𝑔 of π‘₯ is the reflection of this in the 𝑦-axis. Reflecting each of the points of 𝑓 of π‘₯ in the 𝑦-axis gives us the coordinates negative one, one; negative two, two; negative three, three; and negative four, four. The π‘₯-coordinates change sign and the 𝑦-coordinates remain the same. 𝑔 of π‘₯ has π‘₯-values equal to negative one, negative two, negative three, and negative four and corresponding values of 𝑔 of π‘₯ or 𝑦 of one, two, three, and four. From the five options originally listed, this was option (D).

We will now summarize the key points from this video. If we consider any function 𝑦 is equal to 𝑓 of π‘₯, then a reflection in the π‘₯-axis is represented by negative 𝑓 of π‘₯ and a reflection in the 𝑦-axis is represented by 𝑓 of negative π‘₯. Reflecting in the π‘₯-axis does not change the roots of a function, but it does change the sign of the 𝑦-intercept. The opposite is true when reflecting in the 𝑦-axis. This changes the sign of the roots but does not change the 𝑦-intercept. Whilst it was not covered in this video, a combined reflection in the π‘₯-axis and then the 𝑦-axis or vice versa is represented as 𝑦 equals negative 𝑓 of negative π‘₯. It is also worth noting that when reflecting in lines other than the π‘₯- or 𝑦-axis, other approaches are needed and the algebraic interpretation is not so straightforward.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.