quadtratic equation y=x2-5x find co-ordinates of a and b

1 ANSWERS

1. amomipais82
   

   Hi,
   If y= 3x2+4x+ 7, rewrite it as y= 3(x2+ (4/3)x)+ 7
   A "perfect square" is of the form (x+ a)2= x2+ ax+ a2. Compare that to x2+ (4/3)x. The coefficient of x is 4/3. 2a= 4/3 if a= 2/3 and, in that case, a2= 4/9. If we add 4/9 then x2+ (4/3)x+ 4/9= (x+ 2/3)2, a perfect square.
   
   Of course, we can't JUST add 4/9, that would change the value. We can, however both add and subtract 4/9:
   
   y= 3(x2+ (4/3)x+ 4/9- 4/9)+ 7 and now take the "-4/9" out of the parentheses:
   = 3(x2+ (4/3)x+ 4/9)- 4/3+ 7
   = 3(x+ 2/3)2+ 17/3
   
   If x= -2/3, then x+ 2/3= 0 so y= 17/3. If x is ANY OTHER number, x+ 2/3 is NOT 0 so the square is positive and y is 17/3 PLUS some number: larger than 17/3.
   
   The lowest point on the parabola, the vertex, is (-2/3, 17/3).
   
   To "transform a straight line left or right on the x-axis", in other words, translate left or right, ADD OR SUBTRACT SOMETHING FROM x.
   
   If the equation of the line is y= 5x+ 3, then y= 5(x- 2)+ 3 = 5x- 10+ 3= 5x- 7 has graph a line parallel to y= 5x+ 3 but shifted right 2 places: y= 3 in 5x+ 3 when x= 0 but in y= 5x- 7 when x= 2.
   
   The graph of any function, y= f(x), is shifted to the right a places is x is replaced by x-a: y= f(x-a).
   
   Thanks

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