# finding k so line is tangent to graph

• Jan 11th 2009, 10:07 PM
Neverending
finding k so line is tangent to graph
Find k such that the line is tangent to the graph of the function

http://i146.photobucket.com/albums/r...kitty216/4.jpg

don't know how to do this one
• Jan 11th 2009, 10:16 PM
Prove It
Quote:

Originally Posted by Neverending
Find k such that the line is tangent to the graph of the function

http://i146.photobucket.com/albums/r...kitty216/4.jpg

don't know how to do this one

At what value of x does the line need to be tangent to the function?
• Jan 11th 2009, 10:17 PM
mr fantastic
Quote:

Originally Posted by Neverending
Find k such that the line is tangent to the graph of the function

http://i146.photobucket.com/albums/r...kitty216/4.jpg

don't know how to do this one

A non-calculus approach is to solve them simultanously for x and set the discriminant of the resulting quadratic equation equal to zero (so that there's only one intersection point):

$\displaystyle x^2 - kx = 4x - 9 \, ...$
• Jan 11th 2009, 10:33 PM
Neverending
Quote:

Originally Posted by Prove It
At what value of x does the line need to be tangent to the function?

I think it wants all values of x where it would be tangent. It must have to do with the derivative because that's the chapter im on. I just don't know what to do with it.

well the deriv = 2x - k

Am I suppose to set that equal to the slope perhaps?? but when I do I get something like 2x-4... what to do with that?
• Jan 12th 2009, 01:13 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
A non-calculus approach is to solve them simultanously for x and set the discriminant of the resulting quadratic equation equal to zero (so that there's only one intersection point):

$\displaystyle x^2 - kx = 4x - 9 \, ...$

$\displaystyle \Rightarrow x^2 - (k+4)x + 9 = 0$.

Discriminant: $\displaystyle \Delta = (k+4)^2 - 36$.

Solve $\displaystyle \Delta = 0$: $\displaystyle k + 4 = \pm 6 \Rightarrow k = -10, \, 2$.

Calculus approach:

Gradient of tangent = 4 therefore require $\displaystyle \frac{dy}{dx} = 4$:

$\displaystyle 2x - k = 4$ .... (1)

At the point of tangency require the y-coordinate to be the same for the tangent line and the curve:

$\displaystyle x^2 - kx = 4x - 9 \Rightarrow x^2 - (k + 4)x + 9 = 0$ .... (2)

Solve equations (1) and (2) simultaneously for $\displaystyle k$: $\displaystyle k = -10, \, 2$.
• Jan 12th 2009, 11:12 AM
Neverending
yay I got it! Thanks so much (Happy)