i need help this is so hard?

Question

determine a quadratic function F(x)=ax^2 + bx +c if it admits a maximum equal to 9 and passes through the points (1,-7) and (-1,-27)

if you would be kind enought to show the steps so that i can understand that would be really nice thank you

Answer 1

first of all a max is when the derivative is = to 0
F'= 2ax+b=0
F'(9)=2a*9+b=0
18a+b=0


passes by (1,-7) you replace f(x) by -7 and x by 1
-7= a * 1^2 +b*1 +c
a+b+c=-7


passes by (-1,-27) you replace f(x) by -27 and x by -1
-27=a*(-1)^2 + (-1)*b +c
a-b+c=-27

u now have a system of 3 equations and 3 unknowns
18a+b=0 (1)
a+b+c=-7 (2)
a-b+c=-27 (3)

from (1) u can take b=-18a
u replace it in (2) a+(-18a)+c=7
u get -16a+c=7
c=7+16a

u replace it in (3)
a-(-18a)+(7+16a)=-27
a+18a+16a=-27-7
35a= -35
a=-1


b=-18a = -18*-1=18

c=7+16a=7+16*-1=-9

u replace it in F(x)=ax^2 + bx +c
F(x)=-x^2 +18x-9

:D good luck plz double check my calculations cz i done them very quickly hope i helped u :D

Answer 2

Sorry, my last answer was completely wrong.
I misread the problem as "the maximum occurs at 9.".
Here's my corrected solution:
Plug in your 2 points:
-7 = a + b + c =
-27 = a - b + c
Subtracting, we get
2b = 20
b = 10.
Now the maximum(or minimum) of ax²+bx+c always
occurs when x = -b/2a. Also, if a >0 the minimum
occurs at that point and if a < 0 the curve has
a maximum at that point.
You can get this either by completing the square
or by setting the derivative to 0 and solving for x.
So we have
9 = a(b²/4a²) -b(b/2a) + c.
which simplifies to
9 = (b²-4ac)/4a
b²-4ac = -36a
Now b = 10, so
4ac - 36a = 100
a(c-9) = 25.
Finally a+b+c = -7 and b = 10, so
a+ c = -17
a(-26-a) = 25
a²+26a+25 = 0.
(a+1)(a+25) = 0
a = -1, c = -16
or
a = -25, c = 8.
So there are 2 parabolas that solve this problem:
y = -x² + 10x -16
and
y = -25x² + 10x + 8.
Finally, since a < 0 in each case, the function
attains its maximum at 9.

Answer 3

The answer:

This is an upside down parabola along the lines of...

f(x)=-a(x-h)^2+9 with a>0

with (h,9) being the vertex of the parabola.
so
-7=-a(1-h)^2+9
a(1-2h+h^2)=16
ah^2-2ah+a=16

-27=-a(-1-h)^2+9
a(1+2h+h^2)=36
ah^2+2ah+a=36

Subtract and get
4ah=20
ah=5
h=5/a

f(x)=-a(x-5/a)^2+9
f(x)=-ax^2+10x-25/(a)+9

This is the general solution.

Let's finish it...
set -a(1^2)+10-25/a+9=-7
-a^2+10x-25+16a=0
a^2-26x+25=0
(a-25)(a-1)=0
a=1 a=25

Two answers.
f(x)=-x^2+10x-25+9=-x^2+10x-16
f(x)=-25x^2+10x-1+9=-25x^2+10x+8

When I finally figured it out, I felt stupid.
I got the right answer another way but it was long and hard.
*********************************************************************
Some stuff I posted before:


abou.el.leil has done something wrong... because...

F(x)=-x^2 +18x-9
gives F(1)=-1+18-9=8 which is not equal to -7.

I'm looking through his stuff now to try to figure out where he has gone wrong.

If you subtract these two equations:
a+b+c=-7 (2)
a-b+c=-27 (3)

you get -2b=-20 so b=10

But he has b as 18.

Here's where he makes his mistake:
from (1) u can take b=-18a
u replace it in (2) a+(-18a)+c=7
u get -16a+c=7
c=7+16a

He's got two problems here. First he uses 2*a where 2 was the number of the equation and not something to multiply by a.
Second he has =7 instead of = -7 as it's suppose to be.

Also he uses f'(9)=2a(9)+b
This is completely wrong...should be...
Ack I messed up again.
So the answer below me is wrong.
f'(x) is not equal to f(x) so you can't say
f'(x)=9=2ax+b
and you cannot say
f'(9)=0=2a9=b

There is a way to use the derivitive but it is long and hard, you get the right answer in the end. But the way above is easier.
Mutter...

Answer 4

i can't show the steps because it is extremely long. but ill be happy to tell u in email. also i dont get the part after the equation if it admits.........if u cud explain this by email, ill try to answer. 10q.