Originally Posted by
Jhevon recall that the derivative gives the slope of the tangent line to a curve at some point.
now, the derivative here is $\displaystyle y' = 3x^2$ we need to show that there are three such lines with this slope touching our curve.
now, any point on our curve is given by: $\displaystyle \left( x, x^3 \right)$
thus, using $\displaystyle \frac {y_2 - y_1}{x_2 - x_1}$, the slope of any line touching our curve and passing through the point (2,3) is given by:
$\displaystyle \frac {3 - x^3}{2 - x}$
now, if this slope is equal to the slope of our tangent line, we will have a tangent for every x-value given by that formula. thus we want to solve for all x's such that:
$\displaystyle \frac {3 - x^3}{2 - x} = 3x^2$
i leave the rest to you. showing that we have 3 solutions to this equation shows that there will be 3 tangent lines to our curve that pass through (2,3)