Help please :)
Find the values of a and b if the parabola y=a(x+b)^2 −8 has tangent y = 2x at the point P(4, 8),

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- AonZ

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- AonZ

i have gotten gradient to be 2a(x+b)

- myininaya

ok great so far
y' at (x,y)=(4,8) is 2a(4+b)=8a+2b
And we are also given at that point we have y'=2x
But that point is (4,8) so at that point we have y'=8
So this means we have 8=8a+2b
Did you get this far?

- myininaya

We should be looking for another liner equation

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- myininaya

Any ideas?

- AonZ

not really

- AonZ

i got 8=8a +2ab

- myininaya

Well you also know (4,8) is on the parabola. See if you can use that.

- myininaya

this should give you two equations with the only unknowns a and b

- myininaya

You know how to solve a system of equations, correct?

- AonZ

err yea

- myininaya

Did you think to do 8=a(4+b)^2-8

- AonZ

i got 8=8a +2ab not 8=8a +2b btw

- myininaya

wait how did you get the first equation?

- myininaya

oh wait i messed up when i wrote it

- myininaya

2a(4+b)=8a+2b
I didn't distribute there correctly
8a+2ab would be correct

- myininaya

so yeah we have 8a+2ab=8

- AonZ

2nd equation is 16a + 8ba + ab^2 - 16 right?

- myininaya

all of that = 0 yes

- AonZ

then we just solve right?

- myininaya

yes

- AonZ

16a + 8ba + ab^2 - 16 = 0 and 8a+2ab-8= 0
err how?

- myininaya

Did you try solving the linear equation either for a or b
and replace the a or b in the first equation you ahve there

- AonZ

im pretty sure we made a mistake somewhere

- AonZ

http://www.wolframalpha.com/input/?i=16a+%2B+8ba+%2B+ab%5E2+-+16+%3D+0++and++8a-2ab-8%3D+0

- AonZ

the answers in the back of my book says 1/16 = a and 12= b

- myininaya

why did you change the sign in that one equation?

- AonZ

oh typed it wrong before

- AonZ

oh wait...

- AonZ

can we start from the begining again?

- AonZ

the derivative is 8a+2ab
that means the gradient is 8a + 2ab
How did it become
8= 8a + 2ab

- myininaya

y'=2x at (4,8)

- myininaya

we also know that y'=2a(x+b) at (4,8)
so at (4,8) we know 2x=2a(x+b)

- myininaya

x=4 then

- myininaya

so we have 2*4=2a(4+b)

- myininaya

8=8a+2ab

- AonZ

ok alright

- myininaya

your book's answer doesn't make sense

- AonZ

hmm

- myininaya

so if a=1/16, b=12
and we have that y'=2x at (4,8) and y'=2a(x+b) at (4,8)
then we have
2*4=2*1/16(4+12)
correct?
is this true?

- myininaya

2*4=8
2/16*(4+12)=1/8*(16)=2
2 does not equal 8

- AonZ

err yea

- myininaya

The equation you put is correct right?
y=a(x+b)^2-8

- myininaya

the point is (4,8) where we have the tangent line y=2x

- AonZ

yes right question

- myininaya

i see nothing wrong with our resulting equations

- AonZ

hmm yea

- AonZ

but i dont think we could sub (4,8) into 16a + 8ba + ab^2 - 16=0

- myininaya

we already did to get that equation

- myininaya

remember y=a(x+b)^2-8
(4,8) is on that parabola

- myininaya

8=a(4+b)^2-8

- myininaya

a(4+b)^2-16=0
a(16+8b+b^2)-16=0
ab^2+8ab+16a-16=0

- myininaya

So we have
ab^2+8ab+16a-16=0
4=4a+ab

- myininaya

Solve that bottom equation for either a or b then plug into the first equation.

- AonZ

a = 1, b=0

- myininaya

By the way I just divided our earlier linear equation by 2 to get 4=4a+ab
4=a(4+b)
4/(4+b)=a
Assuming b does not equal -4.
Then we have 0=4/(4+b)*b^2+8*(4/(4+b))*b+16*(4/(4+b))-16

- myininaya

Multiply both sides by 4+b
0=4b^2+32b+64-16(4+b)

- AonZ

http://au.answers.yahoo.com/question/index?qid=20130720230137AA4cCvd

- myininaya

yep so i did make a mistake there
the slope of 2x is 2
I put 8. haha
So we had 2=8a+2ab was suppose to be that one linear equation
not 8=8a+2ab

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