1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the equation of a parabola with only 2 x-intercepts and the vertex

  1. Jan 29, 2012 #1
    1. Write an equation of the quadratic function f whose graph has x-intercepts 3 and 7 and f(5) = 8.

    2. I really don't know where to go from here :frown:

    3. obviously, is the x-int. are 3 and 7 that means they are points (3,0) and (7,0). Also, f(5) = 8 becomes the point (5,8) which i have found to be the vertex. But, at this point, I don't know how to form a parabolic equation around those three points. Please help me!!! :confused:
     
  2. jcsd
  3. Jan 29, 2012 #2
    What do you know about the roots of a quadratic? You're really almost there. x=5 and x=7 are roots of f(x) and you need f(5)=8.


    Hint:
    Think of:

    [tex]
    (x-a)(x-b)\cdot K =f(x)

    [/tex]

    Can you take it from here?
     
  4. Jan 29, 2012 #3
    i actually have no idea how to move on from there because we haven't learned that formula. I am in CP Pre-calc in high school and the simplest solution would be great. I am not very good at math and this is a review of stuff we learned from september :/
     
  5. Jan 29, 2012 #4
    Well if a polynomial f(x) has zeros (let's say 2 in this case, it's a quadratic), then f(x), when factored, could look like: (x-a)(x-b)*K=f(x). We say that a and b are roots of f(x) because they are zeros of the polynomial.

    So you had said that (3,0) and (7,0) were solutions to f(x). Do you now see what to do?

    Edit: Use the two "x-values" you have as the roots of your quadratic, and then you'll need that f(5)=8...So you will need to modify your polynomial to make it work, in other words you will need to multiply it by a factor K.

    Does this help at all?
     
    Last edited: Jan 29, 2012
  6. Jan 29, 2012 #5
    well i tried multiplying x-3 and x-7 to get the equation and i got x^2 - 10x + 21. but, if you plug in 5 as x it does't equal 8. and also if point (5,8) is the vertex that means the parabola would open downwards, which makes the equation negative. I'm still really confused though :/
     
  7. Jan 29, 2012 #6
    You're almost there. Just find a constant to multiply f(x)=x^2 - 10x + 21 so that f(5)=8.
     
  8. Jan 29, 2012 #7
    thanks!!
     
  9. Jan 29, 2012 #8
    so wait now all i have to do is multiply the entire formula by one constant until f(5)=8?
     
  10. Jan 29, 2012 #9
    Well whatever polynomial f(x) has f(5)=8 must pass through (5,8).
     
  11. Jan 29, 2012 #10
    but i still can't seem to figure out how to make X^2 - 10x + 21 pass through (5,8)
     
  12. Jan 29, 2012 #11
    With f(x)=x^2 - 10x + 21, what is f(5)?
     
  13. Jan 29, 2012 #12
    the answer becomes -4
     
  14. Jan 29, 2012 #13
    so should i multiply it out by -2 so that the answer becomes 8?
     
  15. Jan 29, 2012 #14

    eumyang

    User Avatar
    Homework Helper

    Just plug in 5 into
    f(x) = k(x2 - 10x + 21):
    8 = k(5^2 - 10(5) + 21)
    ... and solve for k.

    EDIT: Oops, a little too late. ;)
     
  16. Jan 29, 2012 #15
    You got it!
     
  17. Jan 29, 2012 #16
    THANK YOU SO MUCH! I realized i was making a multiplication error early on in the problem and this is why i could not solve it correctly. You're explanation helped so much! thanks again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding the equation of a parabola with only 2 x-intercepts and the vertex
Loading...