Do you know how to solve this problem? I must find the equation of the tangent, given the value of x?

Question

I am given the equation x(x^2-3), x=2
I have already fond the gradient (m). I found first the derivative F'(x) which is f'(x)= 3x^2-3, and later on I found the F'(2), which gave me 9, so the gradient is 9
Now I must find the equation of the tangent, for the time being I have
Y= mx+b
So
Y= 9x+b <-- I have tried but I can't find the value of B, do you know how to find it?

Thank you very much!
Any help is much appreciated! =)

Answer 1

it is so simple u have a point which is (2,f(2) )
f(2)=x(x^2-3) =2(4-3)=2
so it is (2,2)
so
2=9 * 2 +b
b=2-18= -16
so the equation is
y=9x -16

Answer 2

As I understand you, you have
y = x^3-3x and you want the tangent at x = 2.
Well, the tangent at x = 2 must pass through the same point as the curve does and have the same slope.
so the slope = 3x^2-3 which you have already found (1st derivative). Substitute 2 and get 12-3 = 9 which you found.
Now, find y at x = 2
y = 2^3-6 = 2
so
y = mx + b
y = 9x + b
substitute the point y=2, x=2
2 = 18 + b
b = -16
Scott

Answer 3

The point is (2,2) so as you found the gradient the equation is
y-2=9(x-2) so y= 9x -16

Answer 4

the equation of tangent to a curve is
Y-y=m(X-x)
here y=x^3-3x (given curve)
if x=2 y=2^3-3*2=8-6=2
you know m=9
so eq. is

Y-2=9(X-2)
Y-9X+16=0