Can a cubic function have two tangents at a single point?

Solution 1:

Because sometimes a picture really is worth a thousand words:

Solution 2:

The point $(h,3-h)$ need not be on your cubic. To understand what is going on here consider the following simpler situation. Suppose you have the parabola $y=x^2$ and the point $(0,-1)$. This point is not on your parabola, but there are two tangent lines to the parabola that pass through this point. Namely we have the tangent lines $y=2x-1$ and $y=-2x-1$, which are tangent at $(1,1)$ and $(-1,1)$, respectively. These are two distinct tangent lines that happen to both pass through $(0,-1)$. Hopefully this gives a clearer picture as to what is going on here.

Solution 3:

but I can't see how there could be two distinct tangents at a given point.

Very easy: this point does not belong to the curve. There exist exactly two tangents to the curve, touching the curve at different points, that intersect at this point.

Solution 4:

It's just a case of misreading: Take the unit circle and the point $P = (3, 0)$. You can draw "two tangents from $P$ to the unit circle."

The same deal here: you've got the graph of a cubic, and if you take the right two tangents, they MEET at the point $(h, 3-h)$. I'm almost certain this is what the question is asking.

Solution 5:

There is only one tangent to a real polynomial at a given point on the polynomial. But any two lines in the plane may intersect at a point. The question is asking about a point not on the curve: are there two tangent lines $l_i~ \mathrm{and}~ l_j$ which intersect at the point $(h,~ h-3)$?

Consider the parabola $y=x^2 + 1$. The slope of the tangent line at any point $(w,~ w^2 + 1)$ is $2w$, and therefore the equation of this line is

$$y-(w^2+1) = 2w(x-w) \\ y = 2wx - 2w^2 + w^2 + 1 = 2wx - w^2 + 1$$

Consider the point $(0,-8)$. We want this point to be on our line. So

$$-8 = -w^2 + 1 \\ w^2 = 9 \rightarrow w = \pm 3$$ gives us two points where the tangent lines cross at the point we chose.