Question Video: Transforming Functions Graphically and Algebraically Mathematics

The red graph in the figure has the equation 𝑦 = 𝑓(π‘₯), and the blue graph has the equation 𝑦 = 𝑔(π‘₯). Which of the following will not transform the graph of 𝑓(π‘₯) onto the graph of 𝑔(π‘₯)? [A] A reflection in the π‘₯-axis [B] A reflection in the 𝑦-axis, followed by a horizontal shift of 4 units to the left [C] A horizontal shift of 4 units to the right, followed by a reflection in the 𝑦-axis [D] A vertical shift of 4 units down, followed by a reflection in the π‘₯-axis

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Video Transcript

The red graph in the figure has the equation 𝑦 equals 𝑓 of π‘₯, and the blue graph has the equation 𝑦 equals 𝑔 of π‘₯. Which of the following will not transform the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯? (A) A reflection in the π‘₯-axis. (B) A reflection in the 𝑦-axis followed by a horizontal shift of four units to the left. (C) A horizontal shift of four units to the right followed by a reflection in the 𝑦-axis. (D) A vertical shift of four units down followed by a reflection in the π‘₯-axis.

So we’ve been given two graphs, 𝑦 equals 𝑓 of π‘₯ and 𝑦 equals 𝑔 of π‘₯, and asked to determine which of four sets of transformations will not map the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯. We can answer this question both graphically and algebraically. So let’s consider each of the four options in turn. Option (A) is a reflection in the π‘₯-axis. When we reflect over the π‘₯-axis, points that were above the π‘₯-axis move the same distance below it and vice versa. From the graph, we can see that if we were to reflect the line 𝑦 equals 𝑓 of π‘₯ in the π‘₯-axis, it would produce the line 𝑦 equals 𝑔 of π‘₯. So option (A) does transform the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯.

Now we also want to answer this question algebraically. And in order to do this, we’ll need to find the equations of the lines 𝑦 equals 𝑓 of π‘₯ and 𝑦 equals 𝑔 of π‘₯. We won’t go through the algebra of this in any great detail here. But using the slope–intercept form of the equation of a straight line, 𝑦 equals π‘šπ‘₯ plus 𝑏, and the slope formula, π‘š equals 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one, we can determine that 𝑓 of π‘₯ is equal to two π‘₯ plus four and 𝑔 of π‘₯ is equal to negative two π‘₯ minus four.

We can now consider the algebraic effect of each of these transformations. A reflection in the π‘₯-axis corresponds to negation of the entire function; 𝑓 of π‘₯ is mapped to negative 𝑓 of π‘₯. So if we take the function 𝑓 of π‘₯, which was equal to two π‘₯ plus four, and multiply it by negative one, we obtain negative 𝑓 of π‘₯ is equal to negative two π‘₯ minus four, which is the equation of 𝑔 of π‘₯. So this further confirms that option (A) does transform the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯.

Let’s now consider option (B) a reflection in the 𝑦-axis followed by a horizontal shift of four units to the left. We’ll answer this graphically first. The reflection in the 𝑦-axis first causes points to move to the opposite side of the 𝑦-axis to give the graph shown in orange. We then need to perform a horizontal shift of four units to the left. So every point moves left by this amount. And we can see that all these points do lie on the graph of 𝑦 equals 𝑔 of π‘₯. So we’ve shown graphically that option (B) does transform the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯. Let’s now consider the algebraic approach. Reflection in the 𝑦-axis and a horizontal shift of four units to the left each have a horizontal effect, so they each correspond to a change of variable.

A reflection in the 𝑦-axis corresponds to negating the π‘₯-variable. And a horizontal shift of four units to the left corresponds to swapping the π‘₯-variable for π‘₯ plus four. So if we start with 𝑓 of π‘₯ and apply a reflection in the 𝑦-axis, we obtain two multiplied by negative π‘₯ plus four, which is negative two π‘₯ plus four. If we then perform a horizontal shift four units to the left, we obtain negative two multiplied by π‘₯ plus four plus four, which is negative two π‘₯ minus eight plus four. And this simplifies to negative two π‘₯ minus four. That’s the same as the expression for the function 𝑔 of π‘₯. So this confirms that option (B) does transform the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯.

Next, we’ll consider option (C) a horizontal shift of four units to the right followed by a reflection in the 𝑦-axis. Following a horizontal shift of four units to the right, we have the graph shown in orange. Then to reflect in the 𝑦-axis, each point moves to the opposite side of the 𝑦-axis. And we can see that this does indeed map the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯. Algebraically, a horizontal shift of four units to the right corresponds to changing the variable from π‘₯ to π‘₯ minus four. And a reflection in the 𝑦-axis corresponds to negating the variable. So two π‘₯ plus four becomes two multiplied by π‘₯ minus four plus four, which is two π‘₯ minus four, and then two multiplied by negative π‘₯ minus four, which is negative two π‘₯ minus four, the correct expression for 𝑔 of π‘₯.

So we found that options (A), (B), and (C) do all transform the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯. We’re expecting then that option (D) won’t. But we need to check this. A vertical shift of four units down gives the graph shown in orange. And we do need to be a little bit careful here because on the 𝑦-axis each large square represents two units. A reflection in the π‘₯-axis leads to points that were above the π‘₯-axis, moving below it and vice versa. Connecting these points together, we can see that these two transformations do not map the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯.

Let’s also confirm this algebraically. A vertical shift of four units down corresponds to subtracting four from the entire function, and a reflection in the π‘₯-axis corresponds to negating the entire function. So if we start with two π‘₯ plus four, a vertical shift of four units down leads to two π‘₯ plus four minus four, or simply two π‘₯, and then a reflection in the π‘₯-axis leads to negative two π‘₯. This is not the same as the function 𝑔 of π‘₯. So this confirms that option (D) does not transform the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯. So using a graphical and an algebraic approach, we’ve shown that the set of transformations that does not transform the graph of 𝑓 of π‘₯ onto the graph of 𝑔 of π‘₯ is option (D), a vertical shift of four units down followed by a reflection in the π‘₯-axis.

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