### Video Transcript

Graphs π΄ and π΅ in the diagram are the graphs of square root functions. They are symmetric about the origin. The equation of graph π΄ is π¦ equals one-third root π₯ plus two plus one. Knowing that a point reflection about the origin is equivalent to a reflection in the π₯-axis followed by a reflection in the π¦-axis, find the equation of graph π΅.

In this question, weβre told that graph π΄ maps onto graph π΅ by rotation about the origin and that this is equivalent to a reflection in the π₯-axis followed by a reflection in the π¦-axis. So, letβs remind ourselves of the algebraic manipulation that we can apply to our function that achieves these results. Suppose we have a function π¦ equals π of π₯. The function π¦ equals negative π of π₯ is a reflection of that original function in the π₯-axis. Similarly, the graph of π¦ equals π of negative π₯ is a reflection of π¦ equals π of π₯ in the π¦-axis. So, we see that weβre going to take the equation of our original graph and weβre going to apply both of these transformations.

Weβll define π¦ to be equal to π of π₯ such that π of π₯ is a third root π₯ plus two plus one. This will be mapped onto negative π of π₯ by a reflection in the π₯-axis. So, to achieve that reflection in the π₯-axis, letβs evaluate negative π of π₯. That is found simply by multiplying the entire expression by negative one. So, we get negative a third root π₯ plus two plus one. Distributing negative one across our parentheses, and we find that negative π of π₯ is negative one-third root π₯ plus two minus one. So, weβve achieved our reflection in the π₯-axis. And so, our graph of π¦ equals negative π of π₯ will look a little something like this.

We now indeed see that we need to perform our reflection in the π¦-axis to map this onto graph π΅. Since weβre mapping negative π of π₯ onto this function, we now need to find negative π of negative π₯. This will give us the reflection of the graph of π¦ equals negative π of π₯ in the π¦-axis. And all we need to do here is replace π₯ with negative π₯, so we get negative one-third root negative π₯ plus two minus one. And so, we have the equation of graph π΅. We now replace negative π of negative π₯ with π¦, and we have π¦ equals negative one-third root negative π₯ plus two minus one.

Now, itβs often sensible where possible to check our answer. Here, we can choose a couple of points that lie on curve π΅ and check that they satisfy our equation. Since four small squares represents two units, two small squares represents one unit. So, we see that our graph passes through the point two, negative one. In other words, when π₯ is equal to two, π¦ should be equal to negative one. Substituting π₯ equals two into our equation and we get π¦ equals negative one-third times the square root of negative two plus two minus one, which is negative one. So, this point does indeed satisfy our equation.

Weβll check it with one further coordinate. We see that it passes through approximately zero, negative 1.5. So, we substitute π₯ equals zero into our equation, and we get negative one-third times the square root of negative zero plus two minus one, which correct to two decimal places is negative 1.47. And so, since our graph passes a little bit above negative 1.5 on the π¦-axis, we can deduce that this is likely to be correct also.