# Math Help - finding k so line is tangent to graph

1. ## finding k so line is tangent to graph

Find k such that the line is tangent to the graph of the function

don't know how to do this one

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

don't know how to do this one
At what value of x does the line need to be tangent to the function?

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

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):

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

4. 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?

5. 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):

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

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

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

Calculus approach:

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

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

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

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

Solve equations (1) and (2) simultaneously for $k$: $k = -10, \, 2$.

6. yay I got it! Thanks so much