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anonymous

  • 4 years ago

find all polynomials f(t) of degree < or = to 2 whose graphs run through the points (1,1) and (3,3), such that f '(2) = 3. Use f(t) = ax^2 + bx + c and solve linear systems for a, b, c.

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  1. asnaseer
    • 4 years ago
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    you should be able to create 3 equation for the 3 unknowns a,b and c using the information given.

  2. anonymous
    • 4 years ago
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    1= a (1)^2 + b(1)+c 3= a (3)^2 + b(3)+c 3= 2a(2) + b now solve

  3. anonymous
    • 4 years ago
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    howd you create those though?

  4. anonymous
    • 4 years ago
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    I plug in the given information

  5. anonymous
    • 4 years ago
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    first line a x^2 + b x +c= f(x) given point=(1,1) so I plugged in 1 for x and 1 for f(x)

  6. asnaseer
    • 4 years ago
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    ok, it says the graph runs through the point (1,1). so it runs through the point x=1, y=1. use this to get the 1st equation imaran posted above.

  7. anonymous
    • 4 years ago
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    ok so i got the first 2... the a+b+c=1 and 9a+3b+c=3 but now do u get the third?

  8. anonymous
    • 4 years ago
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    I differentiated ax^2 + bx + c

  9. anonymous
    • 4 years ago
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    to get 2 a x+ b

  10. asnaseer
    • 4 years ago
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    you are given f'(2)=3. so differentiate f(x) to get f'(x) and then plug in the values.

  11. anonymous
    • 4 years ago
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    so uget 4a+b=3?

  12. asnaseer
    • 4 years ago
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    BTW: your question should state f(x) = ... and not f(t) = ... the variable is 'x' not 't'

  13. anonymous
    • 4 years ago
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    yes

  14. anonymous
    • 4 years ago
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    the textbook says t

  15. asnaseer
    • 4 years ago
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    I would suggest the textbook has a typo.

  16. anonymous
    • 4 years ago
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    ok so now when i try to solve for a, b, c... i have a+b+c=1 9a+3b+c=3 4a+b=3 so i do (a+b+c=1) - (9a+3b+c=3) = -8a-2b=-2 so now i have the equations a+b+c=1 -8a-2b=-2 4a+b=3 so i do (4a+b=3) - (a+b+c=1) = 3a-c=2 so now i have the equations 3a-c=2 -8a-2b=-2 4a+b=3 but when trying to solve for a you get 0a= 4... and 0a isnt possible...

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