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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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.

Mathematics
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you should be able to create 3 equation for the 3 unknowns a,b and c using the information given.
1= a (1)^2 + b(1)+c 3= a (3)^2 + b(3)+c 3= 2a(2) + b now solve
howd you create those though?

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Other answers:

I plug in the given information
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)
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.
ok so i got the first 2... the a+b+c=1 and 9a+3b+c=3 but now do u get the third?
I differentiated ax^2 + bx + c
to get 2 a x+ b
you are given f'(2)=3. so differentiate f(x) to get f'(x) and then plug in the values.
so uget 4a+b=3?
BTW: your question should state f(x) = ... and not f(t) = ... the variable is 'x' not 't'
yes
the textbook says t
I would suggest the textbook has a typo.
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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