One of those Question 1's... (1 Viewer)

[stuka]

Hmm maybe its coz im tired, but i just dont see the way to do this leaping out at me.

Find a value for b such that y=12x+b is a tangent to y=x^3

Any ideas?
Thanks.

 

[random-1006]

Hmm maybe its coz im tired, but i just dont see the way to do this leaping out at me.

Find a value for b such that y=12x+b is a tangent to y=x^3

Any ideas?
Thanks.

for it to be a tangent ( well for it to be a line we need a point, and a gradient).

ok, y=x^3, dy/dx=3x^2, we equate this to y=mx+b

3x^2= 12
x=+-2

when x=2, on y=x^3, y= 8, so (2,8)
when x=-2, on y=x^3, y=-8 ( -2, -8)

sub each coordinate into into the line, get an equation in b , and solve

ie 8= 12(2) +b , b= -16
-8= 12(-2) +b, b= 16

b=+-16