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  • bonkysleuth
    bonkysleuth's Avatar
    314 posts since Mar '07
    • 05 Aug `08, 4:30PM

      (i)Express x^2 - 3x + 5 in the form (x-a)^2 + b

       

      (ii) sketch the graph of y = x^2 - 3x + 5

       

      For (i) i got (x-1.5)^2 + 11/4

       

      I encountered some problems doing the second part(graph sketching). from the above equation, we can determine that the minimum point for the graph is (1.5, 2.75). now sub x = 0 into equation and you get y= 5. but when you sub y=0, you cant get any answer cuz it's a maths error. How do i go about drawin gthe graph since I've only had 2 points?

       

       

  • starmoonsun
    starmoonsun's Avatar
    64 posts since Jul '07
    • 05 Aug `08, 5:02PM

      you cant find the other point cuz the graph doesnt cut the x-axis

       

      so you should just draw the min pt and y-intercept without the curve intercepting the x-axis.

      also, note that b^2 - 4ac < 0 , which means it is either always positive or always negative. which in this case, it is always positive.

      hope this helps :)

  • Moderator
    eagle
    eagle's Avatar
    17,973 posts since Aug '01
    • 05 Aug `08, 5:05PM

      Starmoonsun explained it perfectly.

       

      Add on:

      it is always positive because

      (i) the coefficient of x^2 is positive
      (ii) the equation has been simplified to (x-1.5)^2 + 11/4

       

       

      Regards,
      Eagle
      Owner of Strategic Tuition

  • jayh272416
    jayh272416's Avatar
    1,930 posts since Aug '07
    • 05 Aug `08, 5:21PM

      You only need to find the points which cut the axis. In this case since the point that cuts the x axis is non-existent, you dont have to put it in. Just leave the 2 points.

  • bonkysleuth
    bonkysleuth's Avatar
    314 posts since Mar '07
    • 05 Aug `08, 5:39PM
      Originally posted by starmoonsun:

      you cant find the other point cuz the graph doesnt cut the x-axis

       

      so you should just draw the min pt and y-intercept without the curve intercepting the x-axis.

      also, note that b^2 - 4ac < 0 , which means it is either always positive or always negative. which in this case, it is always positive.

      hope this helps :)


      thanks all! may i know when i will know whether it is positive or negative when b^2 - 4ac < 0? i think i learnt this in A maths before cuz it seems so familiar.

  • starmoonsun
    starmoonsun's Avatar
    64 posts since Jul '07
    • 05 Aug `08, 5:51PM

      x^2 - 3x + 5

      let a =1 (coeff of x^2)

      b = -3 (coeff of x)

      c = 5

      b^2-4ac = (-3)^2 -4(1)(5) = -11 < 0

  • Moderator
    eagle
    eagle's Avatar
    17,973 posts since Aug '01
    • 05 Aug `08, 6:24PM
      Originally posted by bonkysleuth:


      thanks all! may i know when i will know whether it is positive or negative when b^2 - 4ac < 0? i think i learnt this in A maths before cuz it seems so familiar.

      Because the roots of the equation is {-b +/- sqrt(b^2-4ac)} / 2a

      if b^2-4ac < 0, you cannot square root it. Hence, there's no real solution for the roots of the equation.

      The graph that does not have a real solution for their roots are graphs that do not touch the x-axis. They can be either all positive, or all negative.

      Whether it is positive or negative can be seen from the coefficient of x^2. If the coefficient is positive, the graph is positive. If it is negative, the graph will be negative.

       

      Regards,
      Eagle
      Owner of Strategic Tuition

       

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