A community for students.
Here's the question you clicked on:
 0 viewing
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.
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.

This Question is Closed

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1you should be able to create 3 equation for the 3 unknowns a,b and c using the information given.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01= a (1)^2 + b(1)+c 3= a (3)^2 + b(3)+c 3= 2a(2) + b now solve

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0howd you create those though?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I plug in the given information

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first 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)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1ok, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so i got the first 2... the a+b+c=1 and 9a+3b+c=3 but now do u get the third?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I differentiated ax^2 + bx + c

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1you are given f'(2)=3. so differentiate f(x) to get f'(x) and then plug in the values.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1BTW: your question should state f(x) = ... and not f(t) = ... the variable is 'x' not 't'

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1I would suggest the textbook has a typo.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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) = 8a2b=2 so now i have the equations a+b+c=1 8a2b=2 4a+b=3 so i do (4a+b=3)  (a+b+c=1) = 3ac=2 so now i have the equations 3ac=2 8a2b=2 4a+b=3 but when trying to solve for a you get 0a= 4... and 0a isnt possible...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.