Question

Given that the straight line y=3x+c is a tangent to the curve y=x^2+9x+k, express k in terms of c.

Answer 1

intersection
y=x^2+9x+k , y=3x+c

x^2+9x+k=3x+c
rearrange
x^2+6x+k-c=0
b^2-4ac=0
a=1 b=6 c=k-c

(6)^2-4(1)(k-c)=0
solve

Answer 2

k=c+18

Answer 3

Sam : Thanks again (:
Khalil : Where did get 18 from? (:

Answer 4

first u should solve y'(x)=3 let a the solution then y(a)=c

Answer 5

the line y=3x+c has slope 3 so it must be tangent to y=x^2 + 9x + k at x=-3:

3 = 2x + 9 ---> x=-3

so equating the two equations:

3x + c = x^2 + 9x + k
3(-3) + c = (-3)^2 + 9(-3) + k
-9 + c = 9 - 27 + k
-9 + c = -18 + k
9 + c = k

Answer 6

i think we should equat the equations: y(-3)=c