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The graph of a polynomial is shown - HSC - SSCE Mathematics Advanced - Question 4 - 2023 - Paper 1
Question 4
The graph of a polynomial is shown. Which row of the table is correct for this polynomial? A. $y = -(x - b)(x - c)^2$ $ b > 0$ $ c < 0$ B. $y = -(x - b)... show full transcript
Worked Solution & Example Answer:The graph of a polynomial is shown - HSC - SSCE Mathematics Advanced - Question 4 - 2023 - Paper 1
Step 1
Equation: $y = -(x - b)(x - c)^2$ $ b > 0$ $ c < 0$
Answer
This polynomial is of degree 3 and has the general form of a cubic function. The negative sign indicates that the graph opens downwards. The presence of suggests that the polynomial touches the x-axis at x = c, making it a local maximum or minimum. Since c < 0, it touches the x-axis to the left of the origin, indicating the polynomial has one real root and one double root. We can eliminate this option as the graph behavior does not match.
Step 2
Equation: $y = -(x - b)(x - c)^2$ $ b < 0$ $ c > 0$
Answer
This option also opens downwards with one real root at x = b (since b < 0) and one double root at x = c (as c > 0). The graph intersects the x-axis at b which is to the left of the origin and touches at c, which is to the right of the origin. This creates a polynomial behavior consistent with the described graph.
Step 3
Equation: $y = -x(x - b)(x - c)$ $ b > 0$ $ c < 0$
Answer
In this case, we have a polynomial that has one real root at x = 0, one real root at x = c (negative), and another at x = b (positive). This graph behavior would lead to the polynomial opening down at 0 while having roots at both sides which does not match the provided graph.
Step 4
Equation: $y = -x(x - b)(x - c)$ $ b < 0$ $ c > 0$
Answer
Here, the polynomial again opens downwards. However, it has roots at b (negative) and c (positive) with the x-intercept at 0. This indicates it descends to the left and rises to the right through the positive root c, which contradicts the graph since it should not intersect at a positive value.