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# In the xy-plane, line L intersects y = x^2 + 2 at how many points?

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In the xy-plane, line L intersects y = x^2 + 2 at how many points?  [#permalink]

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01 Nov 2019, 06:48
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55% (hard)

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57% (01:30) correct 43% (01:50) wrong based on 96 sessions

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In the xy-plane, line L intersects y = x^2 + 2 at how many points?

(1) The slope of Line L is 0
(2) The Line passes through (4,5)

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Re: In the xy-plane, line L intersects y = x^2 + 2 at how many points?  [#permalink]

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01 Nov 2019, 07:07
Bunuel wrote:
In the xy-plane, line L intersects y = x^2 + 2 at how many points?

(1) The slope of Line L is 0
(2) The Line passes through (4,5)

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y = x^2 + 2 is a parabola facing upwards with vertex = (0, 2)

(1) The slope of Line L is 0
--> The line is parallel to x-axis and equation of line is y = k, for any number k
--> Many intersections are possible.

Case 1: If k = 1, Equation of line is y = 1 --> no intersections
Case 2: If k = 2, Equation of line is y = 2 --> 1 intersection
Case 3: If k = 3, Equation of line is y = 3 --> 2 intersections
--> Insufficient

(2) The Line passes through (4,5)
Since we do not know the slope of the line, we cannot say exact number of intersections --> Insufficient

Combining (1) & (2),
slope of line is 0 and passes through (4, 5)
--> Equation of line is y = 5
--> Number of intersections = 2 --> Sufficient

IMO Option C
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Re: In the xy-plane, line L intersects y = x^2 + 2 at how many points?  [#permalink]

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01 Nov 2019, 11:01
A line is defined by 1. Slope ( direction ) and 2. a point

Hence C

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Re: In the xy-plane, line L intersects y = x^2 + 2 at how many points?  [#permalink]

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28 Jan 2020, 04:00
Bunuel wrote:
In the xy-plane, line L intersects y = x^2 + 2 at how many points?

(1) The slope of Line L is 0
(2) The Line passes through (4,5)

Parabolas: a line in quadratic form means its a parabola;

Transform y=x^2+2 in the form y=ax^2+bx+c
a=1 (upwards parabola), b=0, c=2

If a > 0 (positive) then the parabola opens upward.
If a < 0 (negative) then the parabola opens downward.

Find the vertex to see where the parabola originates.

Vertex(x)=-b/2a=-(0)/2(1)=0
Vertex(y): y=x^2+2, y=(0)+2=2
Vertex(x,y)=(0,2)

(1) The slope of Line L is 0 insufic

mL=0, then L is a horizontal line, y=mx+b becomes y=0+b, y=b
If y<2, then number of intersections is 0
If y=2, then number of intersections is 1
If y>2, then number of intersections is 2

(2) The Line passes through (4,5) insufic

(1&2) sufic
Line L is y=5, a horizontal parallel to the x-axis;
It intersects 2 points.

Ans (C)
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Re: In the xy-plane, line L intersects y = x^2 + 2 at how many points?  [#permalink]

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28 Jan 2020, 10:14
Bunuel wrote:
In the xy-plane, line L intersects y = x^2 + 2 at how many points?

(1) The slope of Line L is 0
(2) The Line passes through (4,5)

Are You Up For the Challenge: 700 Level Questions

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.
Visit https://www.mathrevolution.com/gmat/lesson for details.

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.

Assume the equation of the line L is $$y=mx+b$$.

Since we have 2 variables ($$m$$ and $$b$$) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

Since we have $$y=x^2+2$$ and $$y=mx+b$$, the number of intersections is the number of roots of the equation $$x^2 + 2 = mx + b$$ or $$x^2 - mx + (2-b) = 0$$.
Then we can determine the number of roots of the equation $$x^2 - mx + (2-b) = 0$$ using its discriminant $$m^2 - 4(2-b)$$.

Since the line passes through $$(4,5)$$, we have $$5 = 4m + b$$ or $$b = 5 - 4m$$.
Then the discriminant is $$m^2 - 4(2-b) = m^2 - 4(2-5+4m) = m^2 -16m + 12$$.

Since the slope of the line is $$0$$, we have $$m = 0$$.
Then the discriminant is $$m^2 - 16m + 12 = 12 > 0$$, which tells the line L, $$y=mx+b$$ and $$y=x^2+2$$ have two intersection.

Since both conditions together yield a unique solution, they are sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: In the xy-plane, line L intersects y = x^2 + 2 at how many points?   [#permalink] 28 Jan 2020, 10:14
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