Question Video: Identifying the Graph of a Logarithmic Function after a Reflection in the 𝑦-Axis | Nagwa Question Video: Identifying the Graph of a Logarithmic Function after a Reflection in the 𝑦-Axis | Nagwa

Question Video: Identifying the Graph of a Logarithmic Function after a Reflection in the 𝑦-Axis Mathematics • Second Year of Secondary School

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The graph of the function 𝑓(π‘₯) = 2 log π‘₯ is reflected symmetrically over the 𝑦-axis to obtain the graph of the function 𝑔(π‘₯). Which of the following is the graph of the function 𝑔(π‘₯)? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

The graph of the function 𝑓 of π‘₯ equals two log π‘₯ is reflected symmetrically over the 𝑦-axis to obtain the graph of the function 𝑔 of π‘₯. Which of the following is the graph of the function 𝑔 of π‘₯?

Now, given all the information we have about our function 𝑔 of π‘₯, there are, in fact, two ways to answer this question. One method is to graph the function 𝑓 of π‘₯ equals two log π‘₯ and then sketch the reflection of this graph in the 𝑦-axis. Alternatively, we can consider the algebraic manipulation that would be required to map 𝑓 of π‘₯ onto 𝑔 of π‘₯. We’re going to choose this latter method in fact and then use the former method to check our working.

Suppose we have some function 𝑦 equals 𝑓 of π‘₯. This is mapped onto the graph of 𝑦 equals 𝑓 of negative π‘₯ by a reflection across the 𝑦-axis. This must mean then that our function 𝑔 of π‘₯ is equal to 𝑓 of negative π‘₯. That’s two log of negative π‘₯. Now that we have an expression for 𝑔 of π‘₯, we can plot the graph using a table of values. At this point, it might be worth reminding ourselves that when we’re dealing with a log, the argument must be greater than zero. This means that the values of π‘₯ we need to substitute into our function 𝑔 of π‘₯ must all be less than zero. This will then mean that the argument negative π‘₯ is positive.

If π‘₯ is negative one, 𝑔 of π‘₯ is two times log of negative negative one or two times log of one. It’s worth noting that log without a base does actually mean log base 10, but the log of one is always zero. So two log of one is also zero. Since one small square represents one unit, we can plot the coordinate negative one, zero on each of our graphs. When we do, we see that the graphs of (A), (B), and (C) do not pass through this point. Then 𝑔 of negative five is two log five, which is 1.40. Two log 10 is 10. And two log 15 to two decimal places is 2.35.

When we plot these remaining three points on the graphs, we see that the only graph that passes through all the points that we’ve plotted is graph (E). So graph (E) is the graph of the function 𝑔 of π‘₯.

Let’s check this by reminding ourselves what the graph of the function two log π‘₯ looks like. We know that the graph of this function must pass through the π‘₯-axis at zero. log base 10 of 10 is one, so two log of 10 is two. And we know that the graph of a logarithmic function has that vertical asymptote. Here, the vertical asymptote is the 𝑦-axis. So there is our function 𝑓 of π‘₯. And we can see that a reflection across the 𝑦-axis does indeed produce the graph (E). So the function 𝑔 of π‘₯ is given by graph (E).

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