# Equation of a Line Tangent

• Jan 15th 2008, 03:49 PM
dumplings
Equation of a Line Tangent
How would you find an equation of the line tangent to f(x) = 2x^2 - 4 at teh point (3, 14) ?

If f(2) = 3 and f(prime)(2) = -1, what would be the tangent line when x = 2?
• Jan 15th 2008, 03:55 PM
Jhevon
Quote:

Originally Posted by dumplings
How would you find an equation of the line tangent to f(x) = 2x^2 - 4 at teh point (3, 14) ?

first find the slope at the point, that is, find $f'(3)$

use that as your $m$ in the point-slope form

the tangent line will be given by: $y - y_1 = m(x - x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point the line passes through, here, this is namely $(3,14)$

Quote:

If f(2) = 3 and f(prime)(2) = -1, what would be the tangent line when x = 2?
here $m = -1$ and $(x_1,y_1) = (2,3)$ (do you see why?)

use the same method as above
• Jan 16th 2008, 04:47 AM
colby2152
Quote:

Originally Posted by dumplings
How would you find an equation of the line tangent to f(x) = 2x^2 - 4 at teh point (3, 14) ?

If f(2) = 3 and f(prime)(2) = -1, what would be the tangent line when x = 2?

Finding an equation tangent to that line at (3, 14) involves finding $f'(3) = 4(3) \rightarrow 12$ The values of the derivative function are the slope values at those points. You have the slope, and the x and y coordinates, so therefore you can solve for the equation of the line:

$y=mx+b$