F(x)= 3x^2 at (2,12)?

Question

tangent line problem

Answer 1

f'(x) = 6x gives the gradient of the tangent.
When x = 2, the gradient is 12.
The tangent at (2,12) is therefore:
y = 12x + c ..........(1)
Since y = 12 when x = 2:
12 = 24 + c
c = - 12.
Substituting this in (1):
y = 12x - 12.

Answer 2

First, find the derivative of F at x=2:

F'(x) = 6x
F'(2) = 12

This gives you the slope of the tangent line. It also passes through the point (2, 12), so using the point-slope form:

y-12 = 12(x-2)
y-12 = 12x - 24
y=12x-12

This is the equation of the tangent line, so we are done.

Answer 3

f(x) = 3x^2
f'(x) = 6x
f'(2) = 6(2) = 12

The gradient or slope of f(x) at (2,12) is 12.

y=mx+c
12=12(2)+c
c = 12 - 24
c = -12

So the equation of the tangent to the line at (2,12) is:
y = 12x - 12

Answer 4

f ` (x) = 6x
f ` (2) = 12 = m
y - b = m.(x - a)
y - 12 = 12.(x - 2)
y = 12x - 12 is equation of tangent that passes thro` (2,12)

Answer 5

y = 3x^2

dy/dx = 6x

the slope ( m ) = 6x 2 = 12

equation of tangent is y - y1 = m(x - x1)
y - 12 = 12( x - 2)

y - 12 = 12x -24
y = 12x -24 + 12

y = 12x -12

Answer 6

f(x)=3x^2
f'(x)=6x, evaluate at x=2
f'(2)=12 (slope)
equation of tangent line: y-12 = 12 (x-2)
y=12x-24+12
y=12x-12
y=12(x-1)