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M24-12-Do graphs y=ax2+b a

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M24-12-Do graphs y=ax2+b a [#permalink] New post 14 May 2013, 21:22
If ac≠0, do graphsy=ax2+b and y=cx2+dintersect?
(1) a=−c

(2) b>d

Hi,
Can anyone show me the underlying concept in this question.
Help will be appreciated.
Thanks
H
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Re: M24-12-Do graphs y=ax2+b a [#permalink] New post 15 May 2013, 00:57
Expert's post
imhimanshu wrote:
If ac≠0, do graphsy=ax2+b and y=cx2+dintersect?
(1) a=−c

(2) b>d

Hi,
Can anyone show me the underlying concept in this question.
Help will be appreciated.
Thanks
H


If ac\neq{0}, do graphs y=ax^2+b and y=cx^2+d intersect?

Notice that graphs of y=ax^2+b and y=cx^2+d are parabolas.

Algebraic approach:

(1) a = -c. Given: y_1= ax^2 + b and y_2=-ax^2 + d. Now, if these two parabolas cross, then for some x, ax^2+b=-ax^2 + d should be true, which means that equation 2ax^2+(b-d)=0 must have a solution, some real root(s), or in other words discriminant of this quadratic equation must be \geq0. So, the question becomes: is discriminant=0-8a(b-d)\geq0? Or, is discriminant=-8a(b-d)\geq0? Now can we determine whether this is true? We know nothing about a, b, and d, hence no. Not sufficient.

(2) b>d. The same steps: if y_1= ax^2 + b and y_2= cx^2 + d cross, then for some x, ax^2 +b=cx^2+d should be true, which means that equation (a-c)x^2+(b-d)=0 must have a solution or in other words discriminant of this quadratic equation must be \geq0. So, the question becomes: is discriminant=0-4(a-c)(b-d)\geq0? Or, is discriminant=-4(a-c)(b-d)\geq0? Now can we determine whether this is true? We know that b-d>0 but what about a-c? Hence no. Not sufficient.

(1)+(2) We have that a=-c and b>d, so y_1= ax^2 + b and y_2=-ax^2 + d. The same steps as above: 2ax^2+(b-d)=0 and the same question remains: is discriminant=-8a(b-d)\geq0 true? b-d>0 but what about a? Not sufficient.

Answer: E.

Else consider two cases.
First case: y=-x^2+1 and y=x^2+0 (upward and downward parabolas). Notice that these parabolas satisfy both statements and they cross each other;

Second case: y=x^2+1 and y=-x^2+0 (also upward and downward parabolas). Notice that these parabolas satisfy both statements and they do not cross each other.

Answer: E.

Also discussed here: m24-12-formatting-error-73363.html

Hope it helps.
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Re: M24-12-Do graphs y=ax2+b a   [#permalink] 15 May 2013, 00:57
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M24-12-Do graphs y=ax2+b a

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