Find k such that the line is tangent to the graph of the function
don't know how to do this one
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?
$\displaystyle \Rightarrow x^2 - (k+4)x + 9 = 0$.
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 $\displaystyle k$: $\displaystyle k = -10, \, 2$.