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