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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## The curve shown above is defined by the ordered-pairs (x,y) tagged by: fskilnik@GMATH ##### This topic has 1 expert reply and 1 member reply ### GMAT/MBA Expert ## The curve shown above is defined by the ordered-pairs (x,y) ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult 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 Answer: ____(C)__ _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 fskilnik@GMATH 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. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### Top Member Legendary Member Joined 02 Mar 2018 Posted: 1171 messages Followed by: 2 members $$y=f\left(x\right)=Ax^2+2Bx+C$$ 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$$ • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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