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

I have a question regarding this question. The question posed is

if from a point ($h,3−h$) exactly two distinct tangents are drawn to $f(x)=x^3−9x^2−px+q$ find $p$ and $q$

I've been waiting all week for someone smarter than me to answer this question, but no one has, so I have to ask my question.

I have forgotten most of my calculus, but I can't see how there could be two distinct tangents at a given point. I know the first derivative gives the slope of the tangent line at a given point. If two tangents are distinct, then they must have different slopes (if they are at the same point), otherwise the lines will be parallel. But the first derivative only gives one slope. For two lines with the same slope to be distinct, they must be parallel. Thus the lines do not go through the same point. But the problem says they do.

In my mind, then, there can't be two tangents at a single point because of this apparent contradiction. I have searched the web and nowhere could I find an example (even a strange, unique condition) where there are two tangents at a point. A function is either differentiable at a point, which means there is one tangent line, or the function is not differentiable, which means there can be any number of tangents.

Further confusing me is this business with $(h, 3-h)$. Does this present some special situation where dual tangents are possible?

I appreciate any light that you can shed on the subject.

• The question is asking for tangents from the point, not at the point. Presumably the point does not lie on the curve itself, so there can be multiple tangents from it, the way you can draw two tangents to a circle from a point outside the circle.
– user856
Commented Apr 28, 2017 at 23:14
• Draw tangents to a cubic curve at any two points where the slopes are not equal. Those tangents are not parallel, so they will meet at some point. Commented Apr 28, 2017 at 23:30
• This is a good example of why you need to pay attention to every single word in a math problem. Commented Apr 29, 2017 at 15:39
• The answer to the question in the title is simply "no, because the function is differentiable." But the title doesn't match the actual question. Commented Apr 29, 2017 at 22:08
• Take a curve with any two non-parallel tangents. Where the tangents intersect is then a point where at least two distinct tangents can be drawn to the curve Commented Apr 29, 2017 at 23:03

Because sometimes a picture really is worth a thousand words:

• @scott - I made this a community wiki post simply because I thought a picture would help illustrate the several good answers that were already present. You should pick your favorite of the other answers and mark it as correct. Commented Apr 29, 2017 at 17:03
• Just because it's a saying doesn't mean it's true. I didn't understand how this picture spoke to the question until I read some of the other answers. So it wasn't worth any words to me without an explanatory caption of some sort. Commented Apr 30, 2017 at 3:33
• @ToddWilcox - I've already said I posted it to illustrate what the other answers were talking about.(as a community wiki post, I get no credit for the upvotes). I didn't explain it exactly because I was illustrating their posts, not making an independent answer. Commented Apr 30, 2017 at 3:47

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.

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.

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.

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.

Everything posted so far clarifies the question but does not answer it.

We start from a better phrasing

If it is always possible to draw, through any point P:$(x,y)=(3-h,h)$, two distinct lines tangent to the cubic C:$y=x^3-9x^2-px+q$, find $p$ and $q$.

SOLUTION. Consider a cubic C with its inflexion point at the origin, $y= x^3+cx$, with tangent T through the origin $y=cx$. Then if $c\ge 0$ there always exists only one line tangent to $C$ through a given point P not on C. If $c<0$ there are two tangents through any point P located between C and T or on T. Through any other point there is only one tangent. Note that there is only one straight line from every point of which two tangents are always possible, and that is T itself. This is evident from inspection of this graph; an analytical proof is left to the reader.

In the question, the inflexion point is at $x=3$ and the tangent through the inflexion point is $$y=y(3)+y'(3)(x-3)=q+27-(p+27)x$$ and if this is to coincide with the line $y=3-x$ on which P always lies, then $p=-26$ and $q=-24$.