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

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

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Video Transcript

Which of the following graphs represents the equation 𝑦 is equal to negative π‘₯ minus one all squared?

In this question, we’re given five graphs, and we need to determine which of these five graphs represents the equation 𝑦 is equal to negative one times π‘₯ minus one all squared. And there’s a few different ways we could go about this. For example, we could try eliminating options by determining points on the curve. However, this method will only work if we’re given the options. So instead, we’re just going to sketch the curve 𝑦 is equal to negative one times π‘₯ minus one all squared.

And to help us sketch this curve, we need to notice something interesting. The equation we’re given is in vertex form. That’s the form 𝑦 is equal to π‘Ž times π‘₯ minus β„Ž all squared plus π‘˜, where π‘Ž, β„Ž, and π‘˜ are real numbers and π‘Ž is not zero. And we can recall that the values of π‘Ž, β„Ž, and π‘˜ give us useful information about our curve. First, the coordinates of the vertex of our parabola will be β„Ž, π‘˜. This is also sometimes called the turning point.

Let’s determine the values of π‘Ž, β„Ž, and π‘˜ for the equation given to us in the question. First, the coefficient of our parentheses is negative one. So, π‘Ž is negative one. Next, we’re subtracting one from π‘₯. So, our value of β„Ž is one. Finally, we have no constant at the end of our expression. So, the value of π‘˜ is zero. Therefore, if we substitute the value of β„Ž is one and π‘˜ is zero, we get the vertex will have coordinates one, zero. And if we want, we can add the coordinates of the vertices to all four of our options to eliminate options.

We see in option (A) the vertex is not at one, zero. In option (B), the vertex is not at one, zero. And in option (D), the vertex is not at one, zero. So, these three options cannot be graphs of the equation given to us in the question. However, it’s not necessary to use elimination to answer this question. So, let’s continue sketching our graph. Next, we recall the value of π‘Ž gives us information about the shape of our parabola. In particular, if π‘Ž is positive, our parabola opens upwards, and if π‘Ž is negative, our parabola opens downwards. In our case, our value of π‘Ž is negative one. And we can see option (C) opens upwards. So, option (C) cannot be correct. And this is enough to answer our question by elimination; only option (E) can represent the graph of this equation.

However, for due diligence, let’s finish the sketch of our curve. We’ve shown the coordinates of the vertex of this parabola is one, zero And it’s a parabola opening downwards. However, there’s an infinite number of parabolas which open downwards with vertex at coordinates one, zero. So, we should also find the coordinates of one extra point on our curve. We’ll find the coordinates of the 𝑦-intercept, which we can find by substituting π‘₯ is equal to zero into the equation of our curve. We get 𝑦 is equal to negative one times zero minus one all squared, which we can evaluate is negative one. So, the 𝑦-intercept of this curve is negative one, which we can see also agrees with option (E).

Therefore, we were able to show of the five given options only option (E) represents the equation 𝑦 is equal to negative one times π‘₯ minus one all squared.

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