Question

if p(x)=(x-1)(x+k) and if the line tangent to the graph of p at the point (4,p(4)) is parallel to the line
5x-y+6=0, then k=?
a. -1
b. -2
c. 1
d. 0
e. 2

Answer 1

parallel to y = 5 x - 6 so slope = 5
so dp/dx = 5 at x = 4

dp/dx = slope(x) = (x-1)(1) + (x+k) (1)

= x - 1 + 1 + k = x+k
at x = 4, dp/dx = 4+k
so
4+k = 5
k = 1

Answer 2

parallel to y = 5 x - 6 so slope = 5
so dp/dx = 5 at x = 4

dp/dx = slope(x) = (x-1)(1) + (x+k) (1)

= x - 1 + x + k = 2x+k-1
at x = 4, dp/dx = 7+k
so
7+k = 5
k =-2