Question Video: Finding the Equation of a Function When Given the Graph of Its Image and a Description of the Transformations Mathematics

The graph represents a function 𝑔(π‘₯) after a vertical positive shift by 3 units followed by a reflection on the π‘₯-axis. Which of the following represents the original function 𝑓(π‘₯)? [A] 𝑓(π‘₯) = βˆ’(π‘₯ + 2)Β³ βˆ’ 2 [B] 𝑓(π‘₯) = βˆ’3(π‘₯ βˆ’ 1)Β³ βˆ’ 6 [C] 𝑓(π‘₯) = (π‘₯ βˆ’ 1)Β³ + 2 [D] 𝑓(π‘₯) = βˆ’(π‘₯ βˆ’ 4)Β³ βˆ’ 2 [E] 𝑓(π‘₯) = (βˆ’3π‘₯ βˆ’ 1)Β³ + 2

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Video Transcript

The graph represents a function 𝑔 of π‘₯ after a vertical positive shift by three units followed by a reflection on the π‘₯-axis. Which of the following represents the original function 𝑓 of π‘₯? Is it (A) 𝑓 of π‘₯ equals negative π‘₯ plus two cubed minus two? Is it (B) 𝑓 of π‘₯ equals negative three π‘₯ minus one cubed minus six? Is it (C) 𝑓 of π‘₯ equals π‘₯ minus one cubed plus two? (D) 𝑓 of π‘₯ equals negative π‘₯ minus four cubed minus two. Or (E) 𝑓 of π‘₯ equals negative three π‘₯ minus one cubed plus two.

We’re told that 𝑔 of π‘₯ comes from the original function 𝑓 of π‘₯ after applying two transformations. We can recover the original function 𝑓 of π‘₯ from 𝑔 of π‘₯ by reversing the transformations but beginning from the second one. The process will give 𝑓 of π‘₯ as an expression involving 𝑔 of π‘₯. And then we can finish this problem by finding an equation for 𝑔 of π‘₯.

So let’s begin with the second transformation which is a reflection over the π‘₯-axis. And we might remember that given some function β„Ž of π‘₯, the corresponding negative β„Ž of π‘₯ is a reflection of the original graph across the π‘₯-axis. We reverse it by applying the same transformation. So we take 𝑔 of π‘₯ and we map it onto negative 𝑔 of π‘₯.

Next, we’ll reverse the first transformation. That was a vertical positive shift by three units. The opposite to this is a vertical negative shift, in other words, shifting or translating the graph three units down. Now, we achieve that by taking the function β„Ž of π‘₯ and subtracting three from it. Or, in the case of our transformed function negative 𝑔 of π‘₯, we subtract three from that, so we have negative 𝑔 of π‘₯ minus three. And since we’ve reversed two transformations, this must be equal to our original function 𝑓 of π‘₯.

So all we need to do now is find the equation for 𝑔 of π‘₯. We simply look at the shape and we can already see that it must come from the parent function 𝑦 equals π‘₯ cubed. It’s definitely a cubic graph. But of course if the coefficient of π‘₯ cubed is positive, the cubic graph has the opposite shape. In other words, it looks like this graph but reflected across the 𝑦-axis.

So to achieve a reflection across the 𝑦-axis, we make the entire function negative. So 𝑦 equals negative π‘₯ cubed definitely has the correct shape and orientation. But of course, 𝑦 equals π‘₯ cubed and 𝑦 equals negative π‘₯ cubed intersect the 𝑦-axis at zero. In fact, this is also the stationary point of the function or the point of inflection. On our curve, that corresponds to the point with coordinates one, negative five. So we take 𝑦 equals negative π‘₯ cubed and translate it one unit right and five down. Translating it one unit right occurs when we subtract one from the value of π‘₯. And to move it five units down, we simply subtract five from the whole function. So it certainly appears as if we have the equation for our function 𝑔 of π‘₯.

It might be sensible just to check this by substituting a couple of the coordinates in. For instance, when π‘₯ is equal to negative one, 𝑦 is equal to negative negative one minus one cubed minus five, which equals three as we expected. Similarly, when π‘₯ is equal to zero, 𝑦 is equal to negative zero minus one cubed minus four, which is negative four also as we expected. So with some certainty, we can be assured that we have the correct equation for this graph. So 𝑔 of π‘₯ is negative π‘₯ minus one cubed minus five, and that means we can now find an expression for 𝑓 of π‘₯.

By replacing 𝑔 of π‘₯ with this expression in our equation for 𝑓 of π‘₯, we have 𝑓 of π‘₯ is negative negative π‘₯ minus one cubed minus five minus three. And then we distribute the parentheses to get π‘₯ minus one cubed plus five minus three, which is π‘₯ minus one cubed plus two. And we can now look at our options and see that that corresponds to option (C) 𝑓 of π‘₯ is equal to π‘₯ minus one cubed plus two.

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