Find p and q for Tangent to y=(1/3)x^3 at 24x+3y+2=0

[m00c0w]

For the curve y = (1/3)x^3, given that 24x + 3y +2 = 0 is the equation of the tangent to the curve at the point (p,q) find p and q.

I rearranged to get y = -8x - 2/3

So the tangent gradient is -8

I differentiated to get x^2 - 9

Therefore x^2 - 9 = -8 and x = 1 or x = -1.

From here how do I distinguish whether x = 1 or x = -1?

Many thanks.

 

[stunner5000pt]

for x =1
substitute this into the equation of the tangent and the equation of the curve

do the same thing for x = -1
if in any of the two cases you get the y value that matches between the curve and the tangent then that point (x,y) lies on the curve, thus the line is tangent to that point on the curve