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Question Video: Identifying Graphs of Quadratic Equations in Vertex Form Mathematics

Which of the following graphs represents the equation 𝑓(π‘₯) = βˆ’(π‘₯ βˆ’ 1)Β²? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

02:28

Video Transcript

Which of the following graphs represents the equation 𝑓 of π‘₯ equals negative π‘₯ minus one squared?

And there are five different graphs to choose from.

So, we’re given a function equation, and we need to identify the correct graph for this equation. There are actually two ways we can answer this. We could draw a table of values and plot the relevant graph from that. Alternatively, we can use what we know about the graph of the most basic quadratic equation alongside knowledge of function transformations to identify the correct graph. Let’s use this method.

Consider the function equation 𝑔 of π‘₯ equals π‘₯ squared. The graph of this equation is a symmetric parabola with a turning point at zero, zero and a vertical line of symmetry given by the 𝑦-axis. We can see that the equation we have, 𝑓 of π‘₯ equals negative π‘₯ minus one squared, looks a little like this. In fact, this is equivalent to taking the original equation, 𝑔 of π‘₯ equals π‘₯ squared, subtracting one from the value of π‘₯, and then multiplying the entire expression by negative one. Hence, we can say that 𝑓 of π‘₯ equals negative 𝑔 of π‘₯ minus one.

So, how does this help? Well, we know that given a general function 𝑦 equals 𝑓 of π‘₯, we can map that onto 𝑦 equals 𝑓 of π‘₯ plus π‘Ž by a translation negative π‘Ž, zero, in other words, a translation π‘Ž units left. Similarly, given a general function 𝑦 equals 𝑓 of π‘₯, we can map that onto 𝑦 equals negative 𝑓 of π‘₯ by reflecting in the π‘₯-axis. This means we map 𝑔 of π‘₯ onto 𝑓 of π‘₯ by translating one unit right and reflecting in the π‘₯-axis.

Notice that the order in which we perform these actions here doesn’t matter; we’ll achieve the same result either way. If this was not the case, we would need to perform a horizontal translation before a reflection of any kind. That gives us this graph. We can see that that corresponds to option (E) here. The graph that represents the equation 𝑓 of π‘₯ equals negative π‘₯ minus one squared is (E).

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