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.

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.htmlHope it helps.

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