Find k such that the line [y=5x-4] is tangent to the function : x^2-kx

Answer 1

Where do the graphs meet?
x^2-kx = 5x-4
x^2 - (k+5)x + 4 = 0
We want there to be a single root, so the discriminant must be zero
(k+5)^2 - 16 = 0
k = -5±4 = -9,-1

check:
f(x) = x^2 - kx
f'(x) = 2x-k

k = -9
x^2+9x = 5x-4
x = -2
slope at (-2,-14) = 5

k = -1
x^2+x = 5x-4
x = 2
slope at (2,2) = 5

Answer 2

f(x) = x^2 - kx
f ' (x) = 2x - k

the slope of the line y = 5x-4 is 5

so 2x-k = 5
k = 2x-5

also we want x^2 - kx = 5x-4
then: x^2 - (2x-5)(x) = 5x-4
x^2 - 2x^2 + 5x = 5x - 4
-x^2 = -4
x = ±2

then k = 4-5 = -1
or
k = -4-5 = -9

check:
case1 , if k = -1
then: x^2 + x = 5x-4
x^2 - 4x + 4 = 0
(x-2)^2 = 0
x-2 = 0
x = 2 ----> only one solution ---> only one point of contac
--> must be tangent when x=2

when k = -9
x^2 + 9x = 5x-4
x^2 + 4x + 4 = 0
(x+2)^2 = 0
x+2 = 0
x = -2 , ----> only one solution ---> only one point of contac
--> must be tangent when x=-2