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The curve shown above is defined by the ordered-pairs (x,y)

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The curve shown above is defined by the ordered-pairs (x,y)

by fskilnik@GMATH » Thu Feb 21, 2019 12:46 pm

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GMATH practice exercise (Quant Class 13)

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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

Answer: [spoiler]____(C)__[/spoiler]
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The curve shown above is defined by the ordered-pairs (x,y)

by fskilnik@GMATH » Fri Feb 22, 2019 5:13 am
fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 13)

Image

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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by deloitte247 » Sun Feb 24, 2019 10:56 am
$$y=f\left(x\right)=Ax^2+2Bx+C$$
A, B and C are given constant
The graph is facing upward, therefore
$$coefficient\ of\ x^{2\ }is\ positive\ $$ A is positive

The graph takes minimum value of 0 for
$$x=-\frac{\left(2B\right)}{2A}=Positive$$
Since A is Positive , B must be negative to make the entire expression positive.
The graph intersects y-axis (x=0) at a positive value of y, so when x=0, y is positive
$$A\left(0\right)^2+2B\left(0\right)+C=positive$$
c = positive
Intersection on the x- axis is at a single point of tangency which is positive so the discrimination is 0 ( it has only one root)

Therefore $$\sqrt{2\left(B\right)^2-4AC}=0$$
$$B^2=AC$$
$$I.\ A\ and\ C\ are\ positive\ -True$$
$$II.\ B^2\ is\ greater\ than\ twice\ the\ value\ of\ AC=false$$
$$III.\ \frac{AC}{B\ }is\ negative\ =True$$
$$I\ and\ III=true$$

$$answer\ is\ Option\ C$$
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