Video: US-SAT04S3-Q08-497194670183

The function 𝑦 = β„Ž(π‘₯) is graphed in the given π‘₯𝑦-plane. Which of the following equations could describe β„Ž(π‘₯)? [A] β„Ž(π‘₯) = βˆ’(π‘₯ + 1)Β²(π‘₯ βˆ’ 2)(π‘₯ βˆ’ 1) [B] β„Ž(π‘₯) = (π‘₯ + 1)Β²(π‘₯ βˆ’ 2)(π‘₯ βˆ’ 1) [C] β„Ž(π‘₯) = βˆ’(π‘₯ βˆ’ 1)Β²(π‘₯ + 2)(π‘₯ + 1) [D] β„Ž(π‘₯) = (π‘₯ βˆ’ 1)Β²(π‘₯ + 2)(π‘₯ + 1)

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Video Transcript

The function 𝑦 equals β„Ž of π‘₯ is graphed in the given π‘₯𝑦-plane. Which of the following equations could describe β„Ž of π‘₯? Is it A) β„Ž of π‘₯ equals negative π‘₯ plus one squared multiplied by π‘₯ minus two multiplied by π‘₯ minus one? Is it B) β„Ž of π‘₯ is equal to π‘₯ plus one squared multiplied by π‘₯ minus two multiplied by π‘₯ minus one? Is it C) β„Ž of π‘₯ is equal to negative π‘₯ minus one squared multiplied by π‘₯ plus two multiplied by π‘₯ plus one? Or is it D) β„Ž of π‘₯ is equal to π‘₯ minus one squared multiplied by π‘₯ plus two multiplied by π‘₯ plus one?

We can begin this question by looking at the three points where the graph touches or crosses the π‘₯-axis. These are the roots, or solutions, to the equation. We know that if π‘₯ equals π‘Ž is a root, then π‘₯ minus π‘Ž is a factor. The three points of intersection are negative one, one, and two. Therefore, we have roots at π‘₯ equals negative one, π‘₯ equals one, and π‘₯ equals two.

When the graph touches and doesn’t cross the π‘₯-axis, we have a double, or repeated, root. In this case, we have a double, or repeated, root at π‘₯ equals negative one. This means that π‘₯ plus one all squared is a factor. A double, or repeated, root will have the power two, or squared. The roots π‘₯ equals one gives us a factor π‘₯ minus one. And the root π‘₯ equals two gives us a factor π‘₯ minus two. As the graph crosses the axis at both of these points, these are single roots.

The equation β„Ž of π‘₯ will have brackets π‘₯ plus one all squared, π‘₯ minus one, and π‘₯ minus two. This rules out options C and D. We can see from the graph that the 𝑦 intercept is positive. The graph crosses the 𝑦-axis above the π‘₯-axis. At this point, we know that the π‘₯-coordinate is zero, so we can substitute π‘₯ equals zero into the two remaining options to decide which one is correct.

Substituting zero into option A gives us 𝑦 equals, or β„Ž of π‘₯ equals, negative zero plus one all squared multiplied by zero minus two multiplied by zero minus one. Zero plus one is equal to one. And squaring this also gives us one. Zero minus two is equal to negative two. Zero minus one is equal to negative one. We need to multiply one by negative two by negative one and take the negative of this answer. This is equal to negative two, which means that the graph of π‘Ž would cross the 𝑦-axis at negative two. We can see from our graph that our 𝑦-intercept was positive. Therefore, this cannot be the equation for β„Ž of π‘₯.

Substituting π‘₯ equals nought into option B gives the same calculation without the negative sign at the beginning. This means that instead of getting a 𝑦-value of negative two, we get a 𝑦-value of positive two. As the 𝑦-intercept is now positive, equation B could be the correct answer, the equation that could describe β„Ž of π‘₯ is π‘₯ plus one all squared multiplied by π‘₯ minus two multiplied by π‘₯ minus one. This is because this equation touches the π‘₯-axis at π‘₯ equals negative one, crosses the π‘₯-axis at π‘₯ equals two and π‘₯ equals one, and has a 𝑦-intercept of positive two.

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