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Wondermath

  • 4 years ago

Okay so a graph has x-intercepts -3, -0.5 1 and 2, and it asks you to find the equation of the function, which passes through the point (-1,-6). So I wrote the parent function f(x)=a(x+3)(x+0.5)(x-1)(x-2). However, when i substituted (-1,-6) I got a different value for a as the textbook did, and it's because they used (2x+1) as a factor instead of (x+0.5). So does that mean that I you can't use decimals in the factor?

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  1. amistre64
    • 4 years ago
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    since those are your zeros, multiply them up

  2. amistre64
    • 4 years ago
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    given intercepts at: a,b,c,d, then: (x-a)(x-b)(x-c)(x-d)=0 will give you your basic equation

  3. amistre64
    • 4 years ago
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    then determine that value of that basic for x=-1 and see what you have to scale by to get it to -6

  4. amistre64
    • 4 years ago
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    decimals are fine for factors; but can be written in other ways

  5. amistre64
    • 4 years ago
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    the issue is in the scalar; most likely its a 2 to account for that

  6. Wondermath
    • 4 years ago
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    what do you mean by scalar?

  7. amistre64
    • 4 years ago
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    its that "a" in front you have a basic equation that can be scaled up or down to fine tune it and keep the same intercepts

  8. amistre64
    • 4 years ago
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    x^2 -1 has the same intercepts as 2(x^2-1) but the graph hits a new set of points everywhere else

  9. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=%28x%2B3%29%28x%2B0.5%29%28x-1%29%28x-2%29%2C+x%3D-1 the wolf say the basic equation is good; x=-1 is approx -6

  10. amistre64
    • 4 years ago
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    i see, with the 2x+1 part youve actually doubled it so you have to divide it back out for the "a"

  11. Wondermath
    • 4 years ago
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    yeah that's the problem :(

  12. amistre64
    • 4 years ago
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    at any rate, your setup was fine.

  13. Wondermath
    • 4 years ago
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    i don't know whether i should use x+0.5 or 2x+1

  14. amistre64
    • 4 years ago
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    2(x+.5) = (2x+1) so its really just a matter of what the author of your text has decided to go with

  15. amistre64
    • 4 years ago
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    2x+1 = 0 when x=-.5 x+.5 = 0 when x = -.5

  16. amistre64
    • 4 years ago
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    their choice to double it to begin with accounts for a difference in the scalars you choose for "a"

  17. amistre64
    • 4 years ago
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    but after everything is expanded out, it should be the same

  18. amistre64
    • 4 years ago
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    yours \[x^4+0.5 x^3-7. x^2+2.5 x+3\] theirs \[x^4+x^3/2-7 x^2+(5 x)/2+3\]

  19. Wondermath
    • 4 years ago
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    oh kk thanks!

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