Which lines through the point $(0,3)$ intersect the graph $s=t^3-t$. Need algebraic criterion in terms of coefficients of the equation of the line. Can intersect 3 times, 2 times, 1 time, or never?
$$s=t^3-t$$
Factored form of $s$ is: $$s=t(t+1)(t-1)$$
Derivative of $s$ is: $$s'=3t^2-1$$
Then I use the point-slope formula with the given coordinates, where $m$ is the slope:
$$s-3=m(t-0)$$ $$s=mt+3$$
The problem states up to 3 times, so I think the equation will be in the following form: $$At^3+Bt^2+Ct+D=0$$
Then, I substitute the $s'$ into the point-slope formula line: $$s=(3t^2-1)t+3$$ $$s=3t^3-t+3$$
I graphed $s=t^3-t$ and $s=3t^3-t+3$ on the same graph and seem to intersect at only one point but am unsure if they intersect out to infinity? Is this the correct method?
Since every cubic polynomial has at least one real root, the graph of a cubic polynomial intersects every line in the plane at least once, including vertical lines.
Given $s(t)=t^3-t$ and a line $y=mt+3$, the line will intersect the graph of $S$ twice only at a point where it is tangent to the curve.
The slope of the tangent line to the curve at the point $(t_0,s_0)$ is $m_0=3t_0^2-1$, so the line containing $(0,3)$ which is tangent to the curve will satisfy the point-slope equation of a line
$$ s_0-3=(3t_0^2-1)(t_0-0) $$
And since $s_0=t_0^3-t_0$ this simplifies to $$2t_0^3=-3$$
So the slope $m_0$ which gives the line intersecting the curve only twice is the line with slope
$$ m_0=3(3/2)^{2/3}-1 $$
Lines through $(0,3)$ having slopes less than $m_0$ (and the vertical line) will intersect the curve only once. Lines through $(0,3)$ having slopes greater than $m_0$ will intersect the curve three times.