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Thread: Calculate f(-1)

  1. #1

    Calculate f(-1)

    I need help solving this one problem assigned for homework in my Algebra II HOnors class.

    Let f(x)=ax^2+bx+c such that
    f(0)=f(1)
    f(2)=2f(1)
    f(3)=16
    Calculate f(-1)

    I'm lost, and have no idea where to start. All I've been able to work out is that -a=b because

    f(0)=f(1)

    f(0)=a(0)^2+b(0)+c
    f(0)=c

    f(1)=a(1)^2+b(1)+c
    f(1)=a+b+c

    a+b+c=c
    a+b=0
    -a=b
    Last edited by NoahSilva2; 09-20-2017 at 06:49 PM. Reason: Didn't read forum rules

  2. #2
    Elite Member
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    Quote Originally Posted by NoahSilva2 View Post
    Let f(x)=ax^2+bx+c such that
    f(0)=f(1)
    f(2)=2f(1)
    f(3)=16

    Calculate f(-1)
    What are your thoughts?

    Please share your work with us ...even if you know it is wrong.

    If you are stuck at the beginning tell us and we'll start with the definitions.

    You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

    http://www.freemathhelp.com/forum/announcement.php?f=33
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Quote Originally Posted by Subhotosh Khan View Post
    What are your thoughts?

    Please share your work with us ...even if you know it is wrong.

    If you are stuck at the beginning tell us and we'll start with the definitions.

    You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

    http://www.freemathhelp.com/forum/announcement.php?f=33
    I'm asking for help...I don't have any thoughts. Or work. I need help solving for f(-1). f(0)=f(1), f(2)=2f(1), and f(3)= 16 are GIVEN. Hope that helps?

  4. #4
    Elite Member
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    No, that doesn't help.

    You demonstrated in your edited post that you DO have an idea where to start.

    Now, move on to the next hint and do the same thing. You're almost there.
    Last edited by tkhunny; 09-20-2017 at 07:05 PM.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5
    Quote Originally Posted by tkhunny View Post
    No, that doesn't help.

    You demonstrated in your edited post that you DO have an idea where to start.

    Now, move on to the next hint and do the same thing. You're almost there.
    ​Figured it out, thank you

  6. #6
    Elite Member
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    Good call. Come back, any time.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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