Question

If the graph of the function defined on [-3,3] by f(x)=x^2+ax+b has an absolute minimum at (-1,-3), determine the value of f(1).

Answer 1

complete the square for a quadratic....?

or calculus?

what you learning? :-)

Answer 2

calculus :)

Answer 3

i can't remember how to do tht part though lol

Answer 4

lol!!

Answer 5

small detail

Answer 6

Lol yeah

Answer 7

so we can compute \(f(1)\) like now

but maybe if we figure out what \(f'\) is, it might also add info

Answer 8

So i would take the derivative of that equation?

Answer 9

totally!

Answer 10

sweet so its 2x+a

Answer 11

and it's a min at (-1, -3)

apparently

Answer 12

yup

Answer 13

OK!! lol so you've got that

it's a min, which means something about the 2nd deriv, i reckon

Answer 14

Isn't 2nd deriv for concavity?

Answer 15

i was meandering

it goes through (3,3)

so you have your other equation too

Answer 16

wait how do you know it goes through that point?

Answer 17

Solve
f'[-1]=0
and
f[-1]=3
to find a and b
replace
and find f[1]

Answer 18

Sorry
f[-1]=-3

Answer 19

\( f=x^2+ax+b \)


\(f(1) = 1 + a + b\)

\( f'=2x+a\)

\( f'(-1)=2(-1)+a = 0 \implies a = 2\)

\(\implies f=x^2+2x+b \)

\(\implies 3 =(-3)^2+2(-3)+b \)

\(\implies b = 0 \)

\(\implies f=x^2+2x+b \)

Answer 20

Who said that f(3)=3?

Answer 21

All what we know is f(-1)=-3

Answer 22

oh yes

Answer 23

yes?

Answer 24

If you do what I said above you find
\[
f(x)=x^2+2 x-2\\
f(1)=1
\]

Answer 25

and a would equal 2.
\[f'(x)=2x+a\]
\[f'(-1)=-2+a\]
\[-2+a=0\]
\[a=2\]

Answer 26

how did u find b again? sorry

Answer 27

@eliesaab is not happy yet better at math than I

i reckon we roll that way :-)

Answer 28

Lol no I appreciate both of your help.

Answer 29

nvm i got b

Answer 30

thank you guys!!

Answer 31

@eliesaab
we also are told the slope :-)

Answer 32

as in the min at (-1, -3)

Answer 33

I got the answer, thanks! :D

Answer 34

yes @Thatonegirl_

you don't have to worry!!

good stuff

close it and forget about it