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

As above

H

#### housemouse

##### Guest
plus minus sixteen

#### shinji

##### Is in A State Of Trance
wasn't this just a harder 2unit question? lol

this q wasn't that hard. lol

#### Lizcat

##### Member
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

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

##### New Member
+ 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

##### New Member
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

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

#### Bricnic

##### Lookatmy Member
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

##### New Member
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

#### *Ninny-mole*

##### The Power Is Yours...
hell yeah...that's at least 2/84...

#### pesila

##### New Member
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.

#### dunno04

##### Member
housemouse said:
plus minus sixteen
yeh
plus minus 16
b got 2 values!
=D