### Video Transcript

The red graph in the figure has the equation π¦ equals π of π₯, and the blue graph has the equation π¦ equals π of π₯. Which of the following will not transform the graph of π of π₯ onto the graph of π of π₯? (A) A reflection in the π₯-axis. (B) A reflection in the π¦-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 π¦-axis. (D) A vertical shift of four units down followed by a reflection in the π₯-axis.

So weβve been given two graphs, π¦ equals π of π₯ and π¦ equals π of π₯, and asked to determine which of four sets of transformations will not map the graph of π of π₯ onto the graph of π of π₯. We can answer this question both graphically and algebraically. So letβs 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 π¦ equals π of π₯ in the π₯-axis, it would produce the line π¦ equals π of π₯. So option (A) does transform the graph of π of π₯ onto the graph of π of π₯.

Now we also want to answer this question algebraically. And in order to do this, weβll need to find the equations of the lines π¦ equals π of π₯ and π¦ equals π of π₯. We wonβt go through the algebra of this in any great detail here. But using the slopeβintercept form of the equation of a straight line, π¦ equals ππ₯ plus π, and the slope formula, π equals π¦ two minus π¦ one over π₯ two minus π₯ one, we can determine that π of π₯ is equal to two π₯ plus four and π 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; π of π₯ is mapped to negative π of π₯. So if we take the function π of π₯, which was equal to two π₯ plus four, and multiply it by negative one, we obtain negative π of π₯ is equal to negative two π₯ minus four, which is the equation of π of π₯. So this further confirms that option (A) does transform the graph of π of π₯ onto the graph of π of π₯.

Letβs now consider option (B) a reflection in the π¦-axis followed by a horizontal shift of four units to the left. Weβll answer this graphically first. The reflection in the π¦-axis first causes points to move to the opposite side of the π¦-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 π¦ equals π of π₯. So weβve shown graphically that option (B) does transform the graph of π of π₯ onto the graph of π of π₯. Letβs now consider the algebraic approach. Reflection in the π¦-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 π¦-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 π of π₯ and apply a reflection in the π¦-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. Thatβs the same as the expression for the function π of π₯. So this confirms that option (B) does transform the graph of π of π₯ onto the graph of π of π₯.

Next, weβll consider option (C) a horizontal shift of four units to the right followed by a reflection in the π¦-axis. Following a horizontal shift of four units to the right, we have the graph shown in orange. Then to reflect in the π¦-axis, each point moves to the opposite side of the π¦-axis. And we can see that this does indeed map the graph of π of π₯ onto the graph of π 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 π¦-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 π of π₯.

So we found that options (A), (B), and (C) do all transform the graph of π of π₯ onto the graph of π of π₯. Weβre expecting then that option (D) wonβt. 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 π¦-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 π of π₯ onto the graph of π of π₯.

Letβs 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 π of π₯. So this confirms that option (D) does not transform the graph of π of π₯ onto the graph of π of π₯. So using a graphical and an algebraic approach, weβve shown that the set of transformations that does not transform the graph of π of π₯ onto the graph of π of π₯ is option (D), a vertical shift of four units down followed by a reflection in the π₯-axis.