Equation of Tangent Line Parallel to x+2y-6=0 for f(x)=x^2-x

[sml92]

Homework Statement

Function: f(x)=x^2-x Line: x+2y-6=0

The Attempt at a Solution

Find an equation of the line that is tangent to the graph of f and parallel to the given line.

 

[rootX]

What is the requirement for two lines to be parallel to each other?

 

[RPierre]

First, you must simplify your line for y = mx + b form.

x + 2y - 6 = 0

-> x - 6 = -2y

-> \frac{x-6}{-2} = y

-> y = \frac{-1x}{2} + 3

Next, to find a line tangent to f(x), we take it's derivitive

f'(x) = 2x - 1

Compare the two slopes. 2 and -1/2. These are in fact perpendicular to each other. See what you can do with that.

 

[Dick]

First, you must simplify your line for y = mx + b form.

x + 2y - 6 = 0

-> x - 6 = -2y

-> \frac{x-6}{-2} = y

-> y = \frac{-1x}{2} + 3

Next, to find a line tangent to f(x), we take it's derivitive

f'(x) = 2x - 1

Compare the two slopes. 2 and -1/2. These are in fact perpendicular to each other. See what you can do with that.

I think you should rethink that. I see a line with slope -1/2 and another line with a variable slope. Perhaps you should equate them.

 

[sml92]

thanks guys