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Does the curve h(x)=y=ax^2+b intersects the x-axis?

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Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

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New post Updated on: 17 Feb 2019, 02:17
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A
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C
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E

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Does the curve h(x)=y=ax^2+b intersects the x-axis?

Statement 1: h(4)>0
Statement 2: a>0

Originally posted by DHAR on 30 Jul 2017, 21:54.
Last edited by Bunuel on 17 Feb 2019, 02:17, edited 1 time in total.
Edited the OA.
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Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

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New post 31 Jul 2017, 16:53
2
Hi,

Question :

Does the curve h(x)=y=ax^2+b intersects the x-axis?

Basically its a parabola equation.

For any parabola, ax2 + bx + C = 0, it will intersect the x axis only if discriminant b2-4ac = 0.

it wont intersect x axis if b2- 4ac<0;

Coming to the question:

y = ax2 + b
dicriminant = -4ab ;
now our aim is to check whether -4ab (discriminant in the given equation) is less than, equal or greater than 0.
1) h(4)>0 = > 16a+b>0 = >from this we cannot say about a & b
we dont know whether a and b is + ve or -ve.

So insuff

2) a>0
so we now have a(>0), but we dont know the value of b.

So insuff.


Combining 1 & 2
16a + b > 0
& a>0

therefore we can infer that b >0

so our discriminant -4ab<0

so it dont have any x intercepts.

Suff

C

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Raju
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Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

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New post 31 Jul 2017, 22:43
gvvsnraju@1 wrote:
Hi,

Combining 1 & 2
16a + b > 0
& a>0

therefore we can infer that b >0

so our discriminant -4ab<0


what if a = 1 and b = -2
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Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

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New post 16 Feb 2019, 07:07
I think the answer should be E.

the question asks us to check if -4ab is negative.

(1): 16a + b > 0

16(1) + 2 > 0; 16(1) - 2 > 0; 16(-1) + 17 > 0

SO, Not sufficient

(2) a>0. Nothing about b. So, -4ab can be positive or negative. Not sufficient.

(1) + (2): Still not sufficient since b can be + or -.

Ans: E.

Bunuel, please let us know if this approach is correct.
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Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

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New post 16 Feb 2019, 11:30
chetan2u, Bunuel

Dear experts: Please check if C is the correct answer. I think E should be the answer.

Regards,
Arup
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Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

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New post 16 Feb 2019, 12:07
Statement 1: 16a + b > 0

Both a and b are unknown and thus, insufficient

Statement 2: a > 0

This tells us ax^2 is always positive, however b could be negative. Insufficient

Both 1 and 2 together:

16a + b is positive and that 16a is positive.

B could be positive, no intersection or B could be a large negative number which could give us 1 or 2 intersections. Therefore insufficient.

E
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Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

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New post Updated on: 16 Feb 2019, 19:22
GMATaspirant641 wrote:
I think the answer should be E.

the question asks us to check if -4ab is negative.

(1): 16a + b > 0

16(1) + 2 > 0; 16(1) - 2 > 0; 16(-1) + 17 > 0

SO, Not sufficient

(2) a>0. Nothing about b. So, -4ab can be positive or negative. Not sufficient.

(1) + (2): Still not sufficient since b can be + or -.

Ans: E.

Bunuel, please let us know if this approach is correct.


Hey GMATaspirant641

After combining when we use a>0 in 16a + b > 0, both cases will be valid, b can be <0 or b can be > 0

*Editing it*

Now if -4ab < 0

This means that determinant can be greater than 0 or less than 0

Giving answer as ,rightly mentioned in previous posts, E
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Originally posted by KanishkM on 16 Feb 2019, 12:17.
Last edited by KanishkM on 16 Feb 2019, 19:22, edited 1 time in total.
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Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?  [#permalink]

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New post 16 Feb 2019, 15:52
1
Two posts above suggest that these inequalities:

16a + b > 0
a > 0

imply that b > 0. That is not the case, as you can see by plugging in numbers: we might have a = 1 and b = -1, for example.

If our parabola is y = ax^2 + b, then when a is positive, the parabola (a U shape) opens upwards, and if a is negative, the parabola opens downwards (an upside-down U shape). The value of b is the y-intercept of the parabola. Using both statements here, we know the parabola opens upwards, because a > 0, and we know that when x=4, the parabola is above the x-axis. But that still leaves two possibilities: maybe the parabola has a y-intercept that is above the origin (b > 0), and lies entirely above the x-axis (e.g. in the parabola y = x^2 + 1) or maybe it starts at or below the x-axis (b < 0), crosses the x-axis in one or two places, and becomes positive by the time x = 4 (e.g. in the parabola y = x^2 - 1, which clearly has two x-intercepts at 1 and -1). So the answer is E.

If the source is claiming the answer is C here, then I'd suggest working with more reliable study materials.
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Re: Does the curve h(x)=y=ax^2+b intersects the x-axis?   [#permalink] 16 Feb 2019, 15:52
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