M24-12-Do graphs y=ax2+b a : Retired Discussions [Locked]
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M24-12-Do graphs y=ax2+b a

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14 May 2013, 21:22
If ac≠0, do graphs$$y=ax2+b$$ and $$y=cx2+d$$intersect?
(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]

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15 May 2013, 00:57
imhimanshu wrote:
If ac≠0, do graphs$$y=ax2+b$$ and $$y=cx2+d$$intersect?
(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.

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.

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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