Find the Line Tangent to y=x^3-2 at (0,-18)

[noonan]

Homework Statement

Find the line passing through the point (0,-18) and tangent to the curve y=x^3-2 at some point.

Homework Equations

y=mx+b

The Attempt at a Solution

Well, I know the derivative is 3x^2, and that should be the slope of the line at the point of tangency. I'm just not sure how to find out what point the curve shares with the line! I know that seems like a pretty bad attempt at a solution, I just need a hint about where to start with this one.

 

[SammyS]

The line you want passes passes through a point on the curve y = x³ - 2 , so, as an ordered pair, this point is (x, x³ - 2). The slope of the line is given as
m = 3x², for the same value of x that's in the ordered pair.

Now, we could use the point - slope form of a line, but the point given to us is the y intercept for any line passing through it. Use the slope-intercept form of a line. y = mx + b, where b = -18.

(x³ - 2) = (3x²)(x) - 18 .

Solve this equation for x, to find the x value for a point on the parabola.