Finding three Constants given only relative Max, and y intercept

[Hypnos_16]

Homework Statement

Find constants a, b, and c such that the graph of f(x) = ax2 + bx + c has a relative maximum at (5 , 12) and crosses the the y - axis at (0 , 3).

Homework Equations

Not sure what to use

The Attempt at a Solution

-- i don't have one --

 

[eumyang]

Since you're given the y-intercept (0, 3), you know that when x = 0, f(x) = 3. Plug in these values into
f(x) = ax² + bx + c
and you should be able to find one of the 3 constants.

You were given the hint in your other thread that the derivative at an extrenum is 0. Find the derivative of
f(x) = ax² + bx + c
and plug in some values. Plug in some more values into f(x) and you'll find yourself with 2 equations and 2 unknowns.

 

[Hypnos_16]

so after plugging in (0,3) i got that c = 3.
f'(x) = 2ax + b
after plugging in (5,12) being the max i got that

f'(5) = 2(a)(5) + b
f'(5) = 10a + b

but i don't see where i can really go from there.

 

[eumyang]

so after plugging in (0,3) i got that c = 3.
f'(x) = 2ax + b
after plugging in (5,12) being the max i got that

f'(5) = 2(a)(5) + b
f'(5) = 10a + b

but i don't see where i can really go from there.

You missed this:

You were given the hint in your other thread that the derivative at an extrenum is 0.

So
f'(5) = 10a + b = 0.

You also need to plug in (5, 12) into
f(x) = ax² + bx + c
and you'll have two equations with two unknowns.

 

[Hypnos_16]

So since i know that (5,12) i know that when x is 5 y is 12
soooooo
f'(x) = 2ax + b
f'(x) = 10a + b = 0
b = 10a

f(5) = 25a + 5b + 3
f(5) = 25a + 5(10a) + 3
f(5) = 25a + 50a + 3
f(5) = 75a + 3
12 = 75a + 3
9 = 75a
a = 0.12
?

something like that?
then just fill in what i know to find b

b = 10a
b = 1.2
?

 

[eumyang]

That's the idea, but you have one sign error...

So since i know that (5,12) i know that when x is 5 y is 12
soooooo
f'(x) = 2ax + b
f'(x) = 10a + b = 0
b = 10a

Here. Should be b = -10a.