tytyros Posted April 10, 2008 Share Posted April 10, 2008 In the XY-plane at what 2 points does the graph of y=(x+a)(x+b) intersect the x-axis? (1) (a+b)=1 (2) The graph intersects the Y-axis at (0,-6) OA is C...but I think B...please correct me .... Some theory first y = Ax^2+Bx+C Vertex is given by [ -B/2A, C- (B^2/4A) ] Questions ask about the x intercept which is given by equating the quadratic eqtn to 0. now solve the equation y=x^2+x(a+b)+ab .....corresponds to Ax^2+Bx+C where A=1, B= (a+b)...and C=ab It is very forward that C1 doesnt help to find the answer.... Look at C2... (0,-6) = [ -B/2A, C- (B^2/4A) ] which means B= 0 and C = -6 y = x^2 - 6 and therefore x co-ordinates are +/- SQRT(6) B is Sufficient ????? How come powerprep says answer is C ??? Quote Link to comment Share on other sites More sharing options...
Lhomme Posted April 10, 2008 Share Posted April 10, 2008 It has to be C We know that it is a parabola and it intersects x axis in two points x = -a and x = -b, hence we just need to find points a and b. We also know that the parabola will intersect the y axis in one point. y=(x+a)(x+b) i.e y = x^2 + x(a+b) +ab .......(i) statement1) (a+b)=1...(ii) This is insufficient to get the values of a and b. We need an equation in terms of a and b. statement2) The graph intersects the Y-axis at (0,-6) We know that the x axis is 0 for the y intercept of the parabola. hence substituting x = 0 and y = -6 in equation (i), we get -6 = ab.......(iii)....This is insufficient to obtain a and b. statement (1) and (2) from (ii) and (iii), we can solve for a and b. Hence C is sufficient. Quote Link to comment Share on other sites More sharing options...
Lhomme Posted April 10, 2008 Share Posted April 10, 2008 In the XY-plane at what 2 points does the graph of y=(x+a)(x+b) intersect the x-axis? (1) (a+b)=1 (2) The graph intersects the Y-axis at (0,-6) OA is C...but I think B...please correct me .... Some theory first y = Ax^2+Bx+C Vertex is given by [ -B/2A, C- (B^2/4A) ] Questions ask about the x intercept which is given by equating the quadratic eqtn to 0. now solve the equation y=x^2+x(a+b)+ab .....corresponds to Ax^2+Bx+C where A=1, B= (a+b)...and C=ab It is very forward that C1 doesnt help to find the answer.... Look at C2... (0,-6) = [ -B/2A, C- (B^2/4A) ] which means B= 0 and C = -6 y = x^2 - 6 and therefore x co-ordinates are +/- SQRT(6) I think this is incorrect. for y = -6, x has to be 0. Hence you will have to substitute x = 0 when you substitute y = -6 in the parabola equatioin. B is Sufficient ????? How come powerprep says answer is C ??? See text above Quote Link to comment Share on other sites More sharing options...
tytyros Posted April 10, 2008 Author Share Posted April 10, 2008 got it....x intercepts are -3, and 2...my approach was completely wrong...thanks a ton !! Quote Link to comment Share on other sites More sharing options...
Dynamo Posted April 10, 2008 Share Posted April 10, 2008 The answer is C i think here from y=(x+a)(x+b) = x^2 + bx + ax + ab = x^2 + x(a+b) +ab Statement 1 y = x^2 + x(a+b) +ab we dont know ab so INSUFF... from here... y = x^2 + x +ab Statement 2 substitute the value in the equation... y = x^2 + x +ab -6 = ab y = x^2 + x +ab now when y = 0 0= X^2+X-6 0=(x-2)(x+3)..... Suff Quote Link to comment Share on other sites More sharing options...
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