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Co-Ordinate Geomtery - Parabola!!

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by apoorva.srivastva » Tue Mar 02, 2010 7:43 am
If the graph represented by f(x) = ax^2 + bx + c passes through (-1,0) and (3,0) , is f(5) > 0?

1.) the y-intercept is 3
2.) f(-4) < f(-3)

please explain how to crack this DS...stcuk on St2.

OA is D
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by kevincanspain » Tue Mar 02, 2010 3:11 pm
We simply need to know whether the parabola is convex (hill shaped) or concave (u shaped) i.e. whether a < 0 or a > 0 (this will be the case if and only if a <0)

If it is hill shaped, y decreases as the distance from x to 1 (the x-coordinate of the vertex) increases.
Since the y coordinate corresponding to x= -4 is less than the y coordinate corresponding to x= - 3, this is indeed the case

SUFF

Using algebra,

f(x) = a(x+1)(x -3) for some real number a

f(-4) = a(-3)(-7)= 21a
f( -3) = a(-2)(-5) = 10a

21a < 10a implies than a < 0 and thus f(5) = a(6)(3) < 0

Thus f(5) is definitely not positive SUFF
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by kevincanspain » Tue Mar 02, 2010 3:16 pm
Drawing graphs of yes and no cases makes this much simpler, by the way!
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by apoorva.srivastva » Tue Mar 02, 2010 10:52 pm
kevincanspain wrote:We simply need to know whether the parabola is convex (hill shaped) or concave (u shaped) i.e. whether a < 0 or a > 0 (this will be the case if and only if a <0)

If it is hill shaped, y decreases as the distance from x to 1 (the x-coordinate of the vertex) increases.
Since the y coordinate corresponding to x= -4 is less than the y coordinate corresponding to x= - 3, this is indeed the case

SUFF

Using algebra,

f(x) = a(x+1)(x -3) for some real number a

f(-4) = a(-3)(-7)= 21a
f( -3) = a(-2)(-5) = 10a

21a < 10a implies than a < 0 and thus f(5) = a(6)(3) < 0

Thus f(5) is definitely not positive SUFF
i couldnt understand the funda behind the hill shaped and and u shaped curve..whats the reasoning behind it..is there a theorem related to tit
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by kevincanspain » Wed Mar 03, 2010 2:16 am
Compare the graphs of y=x^2 and y=-x^2
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