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In the xy-plane, does y=a(x-k)^2+p intersect with x-axis?

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In the xy-plane, does y=a(x-k)^2+p intersect with x-axis? [#permalink]

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New post 10 May 2016, 19:03
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E

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In the xy-plane, does y=a(x-k)^2+p intersect with x-axis?
1) a<0
2) p>0

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[Reveal] Spoiler: OA

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Last edited by MathRevolution on 11 May 2016, 20:35, edited 1 time in total.

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Re: In the xy-plane, does y=a(x-k)^2+p intersect with x-axis? [#permalink]

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New post 11 May 2016, 14:24
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MathRevolution wrote:
In the xy-plane, does y=a(x-k)2+p intersect with x-axis?
1) a<0
2) p>0

* A solution will be posted in two days.

Dear MathRevolution,
Just a question of clarification. By
y=a(x-k)2+p
Do you mean
\(y = a(x-k)^2+p\)
Whether the 2 is intended as an exponent is not completely clear.
Mike :-)
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In the xy-plane, does y=a(x-k)^2+p intersect with x-axis? [#permalink]

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New post 13 May 2016, 04:54
In the original condition, there are 3 variables (a,k and p). In order to match the number of equations to the number of variables, we need 3 equations. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that E is the correct answer. Using the condition 1) and the condition 2) at the same time, we get a diagram like the one attached.
If we look at the diagram, the graph passes through (k,p). As p>0, the graph intersects with x-axis. The answer is yes and the condition is sufficient. Hence, the correct answer is C.


- Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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In the xy-plane, does y=a(x-k)^2+p intersect with x-axis? [#permalink]

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New post 21 Oct 2017, 04:32
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Imo Option C
For intersection with x axis
y=0
\(a{(x-k)}^2\)+p=0
\(a{(x-k)}^2\)=\(\frac{-p}{a}\)
We need to the signs of both p&a
From Both statements \(\frac{-p}{a}\)>0
Therefore answer is Option C.
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In the xy-plane, does y=a(x-k)^2+p intersect with x-axis?   [#permalink] 21 Oct 2017, 04:32
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