Q 1 e????? (1 Viewer)

shinji

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wasn't this just a harder 2unit question? lol

this q wasn't that hard. lol
 

Lizcat

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hehe, the question stumped me abit, but then i said.. i;ll get back to it at the end..
you differentiate the y=x^3, and say that 3x^2 has to equal 12.. find x, and sub into the x^3=12x +b, and there u have b!!
16,-16..
 

angelzstorm

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Lizcat said:
hehe, the question stumped me abit, but then i said.. i;ll get back to it at the end..
you differentiate the y=x^3, and say that 3x^2 has to equal 12.. find x, and sub into the x^3=12x +b, and there u have b!!
16,-16..

yepp + - 16...oh good i was rite :)
 

majibow

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+ or - 16 oh my god i stuffed up i got confusesed and said it was just + 16 and not - 16 because i thought it was a porabola god dam it
 

helz_h

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HELL YES haha every question i got right in that test was such a victory, and this happened to be one of them.
hardest test- so hard. so. hard.
 

emma max

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NOOO!!!!! I started to do that, but got confused half way :( i ended up just leaving it...stupid stupid
 

Bricnic

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I thought this was a question involving the discriminant... don't you have to find values of b which make the discriminant 0, so as to give only one solution (and therefore make 12x +b a tangent)? Could be wrong, but this is how I went about it, not sure what answer I wound up with.
 

pesila

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Bricnic said:
I thought this was a question involving the discriminant... don't you have to find values of b which make the discriminant 0, so as to give only one solution (and therefore make 12x +b a tangent)? Could be wrong, but this is how I went about it, not sure what answer I wound up with.
Discriminant applies for a quadratic. This is a cubic, so you could have differentiated to get a quadratic, but that's not doing anything for you. The basic principle here is that a tangent exists where gradients are equal. so we know the gradient of y = 12x + b is 12, so where y = x^3 has gradient 12, then given that you can change your b value, the x-coordinates give you the first piece of the puzzle to get b.

y = x^3
y' = 3 x^2 = 12
x^2 = 4
x = +/- 2
for x = 2, y = 2^3 = 8. for x = -2, y = (-2)^3 = -8
(2,8), (-2,-8)
y = 12x + b y = 12x + b
8 = 24 + b -8 = -24 + b
b = -16 b = 16
 

pesila

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Sorry Brisnik, you could do it by subtracting the functions eg
y = x^3 -12x - b
y' = 3 x^2 - 12 = 0
solve for x = +/- 2
the rest is as I showed before
discriminant of y' still doesn't help much though since
discriminant = 0 + 12 *3 = 36. Therefore the discriminant is fixed, and independent of b and x, therefore useless.
 

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