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At how many points does the parabola y=f(x)=ax^2+bx+c (a≠0

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At how many points does the parabola y=f(x)=ax^2+bx+c (a≠0

by Max@Math Revolution » Thu Jun 13, 2019 10:59 pm

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At how many points does the parabola y=f(x)=ax^2+bx+c (a≠0) intersect the x-axis?

1) x=f(y) intersects the y-axis at two points.
2) y=f(x-1) intersects the x-axis at two points.
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Source: — Data Sufficiency |
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by Max@Math Revolution » Sun Jun 16, 2019 9:57 pm

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Condition 1)
Condition 1) is equivalent to the statement that y=f(x) intersects the x-axis at two points, simply by switching the roles of x and y. This means that y=f(x) intersects the x-axis at two points.
Condition 1) is sufficient since it yields a unique solution.

Condition 2)
The graph of y = f(x-1) is the same as the graph of f(x), moved to the right by one unit. This horizontal movement does not change the number of roots of y. Therefore, y = f(x) also has two roots.
Condition 2) is sufficient since it yields a unique solution.

Therefore, D is the answer.


Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).
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