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# Does the line represented by y=ax^2+bx+c intersect with

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Does the line represented by y=ax^2+bx+c intersect with [#permalink]

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21 Apr 2006, 02:16
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Does the line represented by y=ax^2+bx+c intersect with X-axis?
1)a>0
2)c<0
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Re: Co-Ordiante [#permalink]

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21 Apr 2006, 10:18
The line would intersect with x asis when y becomes 0, ie, when
ax^2 + bx + c = 0 holds true.

From 1, a > 0.
Thus x^2 is not a zero entity. Thus, the line would have a exponential component in the graph.
Also, x^2 would always be positive, and so ax^2 is a positive entity.

But we don't know the signs of b and c and therefore don't know if this expression turns 0.

From 2, c < 0. Thus, for x = 0, the expression is -ve. If the expression is positive for some values of x, then the expression would also be 0 before turning -ve. but we can't assume that, since we don't know the value of a and b.

Combining, we don't know the value of b. But for big enough values of x, ax^2 would be bigger than bx + c and the expression would be +ve.

Therefore, C.

Can someone solve it in a less cumbersome way?
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Re: Co-Ordiante [#permalink]

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21 Apr 2006, 10:29
kapslock wrote:
Can someone solve it in a less cumbersome way?

Just remember the properties:
If a>0 & c>0 parabola opens upwards and the vertex of the parabola is above x axis and the curve will not intersect the x-axis
If a>0 & c <= 0, the parabola opens upwards and the vertex of parabola is touching x-axis or is below the x-axis. Hence it has one or two points.

The opposite is true for a<0... parabola opens downwards....

You can verify this by putting x=0, If c is -ve, kowning that a>0, the parabola should intersect the x-axis at 2 points!
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Re: Co-Ordiante [#permalink]

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21 Apr 2006, 10:40
giddi77 wrote:
kapslock wrote:
Can someone solve it in a less cumbersome way?

Just remember the properties:
If a>0 & c>0 parabola opens upwards and the vertex of the parabola is above x axis and the curve will not intersect the x-axis
If a>0 & c <= 0, the parabola opens upwards and the vertex of parabola is touching x-axis or is below the x-axis. Hence it has one or two points.

The opposite is true for a<0... parabola opens downwards....

You can verify this by putting x=0, If c is -ve, kowning that a>0, the parabola should intersect the x-axis at 2 points!

Girish, one correction.....with a>0 & c>0, the parabola may or may not touch/intersect x-axis

because, possibilities are
b^2 < 4ac,
b^2 = 4ac,
b^2 > 4ac

But with a>0 & c<0, we can definitely say that b^2-4ac > 0, hence intersects x-axis.
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Re: Co-Ordiante [#permalink]

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21 Apr 2006, 11:38
vivek123 wrote:
giddi77 wrote:
kapslock wrote:
Can someone solve it in a less cumbersome way?

Just remember the properties:
If a>0 & c>0 parabola opens upwards and the vertex of the parabola is above x axis and the curve will not intersect the x-axis
If a>0 & c <= 0, the parabola opens upwards and the vertex of parabola is touching x-axis or is below the x-axis. Hence it has one or two points.

The opposite is true for a<0... parabola opens downwards....

You can verify this by putting x=0, If c is -ve, kowning that a>0, the parabola should intersect the x-axis at 2 points!

Girish, one correction.....with a>0 & c>0, the parabola may or may not touch/intersect x-axis

because, possibilities are
b^2 < 4ac,
b^2 = 4ac,
b^2 > 4ac

But with a>0 & c<0, we can definitely say that b^2-4ac > 0, hence intersects x-axis.

Absolutely! I missed that one.
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- Bernard Edmonds

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Re: Co-Ordiante [#permalink]

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21 Apr 2006, 19:53
giddi77 wrote:
vivek123 wrote:
giddi77 wrote:
kapslock wrote:
Can someone solve it in a less cumbersome way?

Just remember the properties:
If a>0 & c>0 parabola opens upwards and the vertex of the parabola is above x axis and the curve will not intersect the x-axis
If a>0 & c <= 0, the parabola opens upwards and the vertex of parabola is touching x-axis or is below the x-axis. Hence it has one or two points.

The opposite is true for a<0... parabola opens downwards....

You can verify this by putting x=0, If c is -ve, kowning that a>0, the parabola should intersect the x-axis at 2 points!

Girish, one correction.....with a>0 & c>0, the parabola may or may not touch/intersect x-axis

because, possibilities are
b^2 < 4ac,
b^2 = 4ac,
b^2 > 4ac

But with a>0 & c<0, we can definitely say that b^2-4ac > 0, hence intersects x-axis.

Absolutely! I missed that one.

I'm sure that was in sleep mode. Dude, can u lend me your maths skills for a day?
Re: Co-Ordiante   [#permalink] 21 Apr 2006, 19:53
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