Expert: Scotto - 4/19/2009

Question
Find the tangent line(s) to the curve y=x^3-9 through the point (1,-9)

Answer
The tangent line needs the same slope as the curve at the point where it touches the curve.  Take the point to be (x0,y0) to be the point where the line hits the curve.  The slope where it will hit the curve is the derivative of the curve, which is 3x @ x0, which is 3x0.

The line would then be y - y0 = 3(x0)²(x-x0).  As can be seen, this line could be a simple one, like y = -9.  That would touch the curve at x=0, for there y would be 0^3 - 9 = -9.  The slope would be the same since the slope of the line is 0 and the slope of the curve is 0.  The x value 1 would be on a point in this line, for y is -9 everywhere.

Putting (1,-9) into the equation for the line will hopefully gives us the other equation -9 - y0 = 3(x0)²(1-x0).   From here we can see that -y0 = 3(x0)² - 3(x0)^3 + 9.  Now since the point (x0,y0) lies on the curve, we know that y0 = (x0)^3 - 9.  Putting this into the equation for the line gives -(x0)^3 + 9 = 3(x0)² - 3(x0)^3 + 9.

The 9's cancels, leaving us with -(x0)^3 = 3(x0)² - 3(x0)^3..  Adding (x0)^3 to both sides gives us
0 = 3(x0)² - 2(x0)^3.  This factors to 0 = (x0)²(3 + 2(x0)).

This says that x0=0, which is the first line I talked about.  It also says that 2(x0) = -3, so x0 = -3/2.  Since y0 = (x0)^3 – 9,
y0 = -27/8 – 72/8 = -99/8.

So the equation of the line is y0 + 99/8 = (27/4)(x + 3/2).
That is the same as y0 + 99/8 = (54x + 81)/8, which is the same as
y0 = (54x –18)/8 = (27x – 9)/8.

Thus, in this response, there are two lines.