How do u find the equation to a tangent line to y=3x^(2)-1 when x=2?

Question

final answer should be y=12x-13

Answer 1

For the line y=3x^(2) - 1 you must first take the derivative of the function. Just use the power rule.
y'= 2*3x^(2-1) - 0 = 6x.
y'=6x will give you the slope of the line at any point x. So if you want to know the equation for the line tangent at x=2 plug in 2 for x.
y'=6*2=12.
You all ready know the x coordinative of the equation, but you have the know the y coordinate before you can find the tangent line, so substitute x=2 into the original equation.
y=3(2)^(2) - 1 = 3(4) - 1 = 11
So now use the standard point-slope form of a line to find the equation of the tangent line since we know the slope and the x and y coordinates.
y - 11 = 12(x - 2)
y - 11 = 12x - 24
y = 12x - 13.
There's your answer. Hope that was helpful.

Answer 2

first find the derivative of the function

f'(x) = 6x

put x = 2
f'(2) = 12

thus the value of gradient of tangent at x = 2 is 12.

find the y coordinate at that point, put x = 2 in the equation y = 3x^2 - 1

y = 3(2)^2 - 1
y = 11

now, you have the point (2,11) of the tangent and the gradient of tangent, m = 12.

use the formula y-y1 = m(x-x1) to form the equation of tangent

y - 11 = 12(x - 2)
y = 12x - 24 + 11
y = 12x -13

that's the answer

Answer 3

plug 2 in the equation to get y. Then you'll have a point on the line.

Find the derivitive y'
plug 2 into the derivitive to get the slope.
Now you have the slope and a point.
Use the point-slope form for the equation of a line.