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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),

Mathematics
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i have gotten gradient to be 2a(x+b)
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?
We should be looking for another liner equation

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Other answers:

Any ideas?
not really
i got 8=8a +2ab
Well you also know (4,8) is on the parabola. See if you can use that.
this should give you two equations with the only unknowns a and b
You know how to solve a system of equations, correct?
err yea
Did you think to do 8=a(4+b)^2-8
i got 8=8a +2ab not 8=8a +2b btw
wait how did you get the first equation?
oh wait i messed up when i wrote it
2a(4+b)=8a+2b I didn't distribute there correctly 8a+2ab would be correct
so yeah we have 8a+2ab=8
2nd equation is 16a + 8ba + ab^2 - 16 right?
all of that = 0 yes
then we just solve right?
yes
16a + 8ba + ab^2 - 16 = 0 and 8a+2ab-8= 0 err how?
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
im pretty sure we made a mistake somewhere
http://www.wolframalpha.com/input/?i=16a+%2B+8ba+%2B+ab%5E2+-+16+%3D+0++and++8a-2ab-8%3D+0
the answers in the back of my book says 1/16 = a and 12= b
why did you change the sign in that one equation?
oh typed it wrong before
oh wait...
can we start from the begining again?
the derivative is 8a+2ab that means the gradient is 8a + 2ab How did it become 8= 8a + 2ab
y'=2x at (4,8)
we also know that y'=2a(x+b) at (4,8) so at (4,8) we know 2x=2a(x+b)
x=4 then
so we have 2*4=2a(4+b)
8=8a+2ab
ok alright
your book's answer doesn't make sense
hmm
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?
2*4=8 2/16*(4+12)=1/8*(16)=2 2 does not equal 8
err yea
The equation you put is correct right? y=a(x+b)^2-8
the point is (4,8) where we have the tangent line y=2x
yes right question
i see nothing wrong with our resulting equations
hmm yea
but i dont think we could sub (4,8) into 16a + 8ba + ab^2 - 16=0
we already did to get that equation
remember y=a(x+b)^2-8 (4,8) is on that parabola
8=a(4+b)^2-8
a(4+b)^2-16=0 a(16+8b+b^2)-16=0 ab^2+8ab+16a-16=0
So we have ab^2+8ab+16a-16=0 4=4a+ab
Solve that bottom equation for either a or b then plug into the first equation.
a = 1, b=0
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
Multiply both sides by 4+b 0=4b^2+32b+64-16(4+b)
http://au.answers.yahoo.com/question/index?qid=20130720230137AA4cCvd
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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