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Functions:using discriminant to find number of x-intercepts

  1. May 21, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-5-21_13-0-5.png

    2. Relevant equations
    Discriminant: b2 - 4ac

    3. The attempt at a solution
    a)
    f(x) = 3x2 – 5x + 1

    y = 3x2 – 5x +1

    Substitute a=3, b= -5, c=1 into the discriminant:

    b2 – 4ac = (-5)2 – 4(3)(1)

    =25 – 12

    = 13 This number is positive.

    Since b2 – 4ac > 0, there are two real roots, so the quadrative function has two x-intercepts.

    b) f(x) = 2x2 + x +1

    y = 2x2 + x +1

    Substitute a=2, b=1, c=1 into the discriminant:

    b2 – 4ac = (1)2 – 4(2)(1)

    = 1 – 8

    = -7 This number is negative.

    Since b2 – 4ac < 0, there are no real roots, so the quadratic function does not have any x-intercepts.

    c) f(x) = 4x2 – 12x + 9

    y = 4x2 – 12x + 9

    Substitute a=4, b= -12, c=9 into the discriminant:

    b2 – 4ac = (-12)2 – 4(4)(9)

    =144 – 144

    = 0

    Since b2 – 4ac = 0, there is one (double) real root, so the quadratic function has one x-intercept.

    Is this correct?
     
  2. jcsd
  3. May 21, 2016 #2
    This is absolutely correct. You can additionally verify by drawing the graphs for the equations.
     
  4. May 21, 2016 #3
    Okay. Thanks for verifying my answer :biggrin:
     
  5. May 21, 2016 #4
    You are welcome. However you could have done it yourself by hand-drawing the graph or using a graphing calculator.
     
  6. May 21, 2016 #5
    Oh okay. But this is the way they taught me in the lesson, they did not ask me to graph anything.
     
  7. May 21, 2016 #6
    What you did / they taught is correct. What I am saying is instead of asking here you could have verified your answer yourself by graphing. It's also a good way to check your answer when you don't have access to the forum.
     
  8. May 21, 2016 #7
    Oh okay, I get it :)
    Thanks again for the help.
     

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