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Re: The curve shown above is defined by the ordered-pairs (x,y) such that [#permalink]
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fskilnik wrote:
GMATH practice exercise (Quant Class 13)



The curve shown above is defined by the ordered-pairs (x,y) such that y = f(x) = Ax^2+2Bx+C, where A, B and C are given constants. If the point of tangency with the x-axis has a positive x-coordinate, which of the following must be true?

I. A and C are both positive.
II. B^2 is greater than twice the value of AC.
III. AC/B is negative.

(A) I only
(B) I and II only
(C) I and III only
(D) All of them
(E) None of them



y = Ax^2 + 2Bx + C

The graph is upward facing. So coefficient of x^2 is positive i.e. A is positive.

Also, the graph takes the minimum value of 0 for x = -(2B)/2A
This value of x is positive.
-B/A is positive. Since A is positive, B must be negative to make the entire expression positive.

Also the graph intersects the y axis (x = 0) at a positive value of y. So when x = 0, y is positive.
A*(0)^2 + 2B*(0) + C = Positive
So C is positive

It intersects the x axis at a single point of tangency which is positive so the discriminant is 0 (It has only 1 root).
So \(\sqrt{(2B)^2 - 4AC} = 0\)
\(B^2 = AC\)

I. A and C are both positive. - True
II. B^2 is greater than twice the value of AC. - False
III. AC/B is negative. - True

Answer (C)
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The curve shown above is defined by the ordered-pairs (x,y) such that [#permalink]
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fskilnik wrote:
GMATH practice exercise (Quant Class 13)



The curve shown above is defined by the ordered-pairs (x,y) such that y = f(x) = Ax^2+2Bx+C, where A, B and C are given constants. If the point of tangency with the x-axis has a positive x-coordinate, which of the following must be true?

I. A and C are both positive.
II. B^2 is greater than twice the value of AC.
III. AC/B is negative.

(A) I only
(B) I and II only
(C) I and III only
(D) All of them
(E) None of them

\(y = A{x^2} + 2Bx + C\)

\(A > 0\,\,:\,\,\,{\rm{parabola}}\,\,{\rm{concave}}\,\,{\rm{upward}}\,\)

\(C > 0\,\,:\,\,\,y - {\rm{intercept}}\,\,{\rm{ > }}\,\,{\rm{0}}\,\,\,\,\,\,\,\,\left[ {f\left( 0 \right) = A \cdot {0^2} + 2B \cdot 0 + C\,\,\, \Rightarrow \,\,\,\left( {0,C} \right) \in {\rm{curve}}} \right]\)

\({\rm{tangency}}\,\,:\,\,0 = \Delta = {\left( {2B} \right)^2} - 4AC = 4\left( {{B^2} - AC} \right)\,\,\,\,\, \Rightarrow \,\,\,{B^2} = AC\)


\({\rm{I}}.\,\,A,C\,\,\mathop > \limits^? \,\,0\,\,\,\left[ {{\rm{True}}} \right]\)

\({\rm{II}}{\rm{.}}\,\,{B^2}\,\,\mathop > \limits^? \,\,2AC\,\,\,\left[ {{\rm{False}}} \right]\,\,\,:\,\,\,{B^2} = AC\,\,\mathop < \limits^{AC\, > \,0} 2AC\)

\({\rm{III}}{\rm{.}}\,\,{{AC} \over B}\,\,\mathop = \limits^{\left( * \right)} \,\,B\,\,\mathop < \limits^? \,\,0\,\,\,\left[ {{\rm{True}}} \right]\,\,\,:\,\,\,0\mathop < \limits^{{\rm{stem!}}} {x_{{\rm{vert}}}} = - {{2B} \over {2A}} = - {B \over A}\,\,\,\,\,\mathop \Rightarrow \limits^{A\, > \,0} \,\,\,\,B < 0\)

\(\left( * \right)\,\,B = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{\\
\,AC = {B^2} = 0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,A\,\,{\rm{or}}\,\,C\,\,{\rm{zero}}\,,\,\,{\rm{impossible}} \hfill \cr \\
\,y = f\left( x \right) = A{x^2} + C\,\,\,\,\, \Rightarrow \,\,\,\,y{\rm{ - axis}}\,\,{\rm{is}}\,\,{\rm{symmetry}}\,\,{\rm{axis}}\,{\rm{,}}\,\,{\rm{impossible}}\,\,\,\left( {{\rm{stem}}} \right) \hfill \cr} \right.\)


The correct answer is (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: The curve shown above is defined by the ordered-pairs (x,y) such that [#permalink]
PriyankGaur wrote:
f(x) has equal and positive roots, hence Discriminant is zero. This gives B^2=A*C. For the type of parabola given, A is positive, hence C is also positive. Therefore I is correct.

From the above point, root would be -B/A and this is given positive. Therefore B is negative. So, III follows.[/color]

From the first point, II is false.

IMO C

Posted from my mobile device



Please explain how b negative. Thank you.
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Re: The curve shown above is defined by the ordered-pairs (x,y) such that [#permalink]
VeritasKarishma wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 13)



The curve shown above is defined by the ordered-pairs (x,y) such that y = f(x) = Ax^2+2Bx+C, where A, B and C are given constants. If the point of tangency with the x-axis has a positive x-coordinate, which of the following must be true?

I. A and C are both positive.
II. B^2 is greater than twice the value of AC.
III. AC/B is negative.

(A) I only
(B) I and II only
(C) I and III only
(D) All of them
(E) None of them



y = Ax^2 + 2Bx + C

The graph is upward facing. So coefficient of x^2 is positive i.e. A is positive.

Also, the graph takes the minimum value of 0 for x = -(2B)/2A
This value of x is positive.
-B/A is positive. Since A is positive, B must be negative to make the entire expression positive.

Also the graph intersects the y axis (x = 0) at a positive value of y. So when x = 0, y is positive.
A*(0)^2 + 2B*(0) + C = Positive
So C is positive

It intersects the x axis at a single point of tangency which is positive so the discriminant is 0 (It has only 1 root).
So \(\sqrt{(2B)^2 - 4AC} = 0\)
\(B^2 = AC\)

I. A and C are both positive. - True
II. B^2 is greater than twice the value of AC. - False
III. AC/B is negative. - True

Answer (C)




Ma'am please explain why b must be negative. Thank you.
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Re: The curve shown above is defined by the ordered-pairs (x,y) such that [#permalink]
saksham1 wrote:
VeritasKarishma wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 13)

The graph is upward facing. So coefficient of x^2 is positive i.e. A is positive.

-B/A is positive. Since A is positive, B must be negative to make the entire expression positive.

Answer (C)


Ma'am please explain why b must be negative. Thank you.


-(B/A)>0, and it was just shown that A>0. But we need the ratio in the brackets to be <0, because only then -(ratio) will be more than zero. So, B/A must be <0 & A>0. Then it must be that B<0.
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Re: The curve shown above is defined by the ordered-pairs (x,y) such that [#permalink]
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