Question

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.

Answer 1

you should be able to create 3 equation for the 3 unknowns a,b and c using the information given.

Answer 2

1= a (1)^2 + b(1)+c
3= a (3)^2 + b(3)+c
3= 2a(2) + b

now solve

Answer 3

howd you create those though?

Answer 4

I plug in the given information

Answer 5

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)

Answer 6

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.

Answer 7

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

Answer 8

I differentiated ax^2 + bx + c

Answer 9

to get 2 a x+ b

Answer 10

you are given f'(2)=3. so differentiate f(x) to get f'(x) and then plug in the values.

Answer 11

so uget 4a+b=3?

Answer 12

BTW: your question should state f(x) = ...
and not f(t) = ...
the variable is 'x' not 't'

Answer 13

yes

Answer 14

the textbook says t

Answer 15

I would suggest the textbook has a typo.

Answer 16

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...