The question reads:

"Suppose that the largest number of points that any horizontal line intersects the graph of a polynomial f(x) is m. Prove that the degree of f(x) is at least m".

I understand the proof graphically - the degree of a polynomial determines the number of local maximum or minimum points minus (-) one (1). If the degree is 4, there are 3 local maximum/minimum points when the function is graphed. And the curves of the graph that make those points allow for a horizontal line to intersect it 4 times at some specific y value, 4 being also the degree of the function. But how does one prove what is asked in the question algebraically? Please help. Thank you very much if you decide to!


closed as off-topic by Marios Gretsas, Claude Leibovici, user91500, Henrik, Xander Henderson Sep 15 '17 at 13:53

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    $\begingroup$ Hint. What's the maximum number of roots that the polynomial $f(x) -a$ can have? You don't need the calculus of turning points, although that's good intuition. $\endgroup$ – Ethan Bolker Sep 14 '17 at 23:49
  • $\begingroup$ What is a? Sorry if it's a bad question $\endgroup$ – Arthur Alex Karapetov Sep 15 '17 at 0:06
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    $\begingroup$ $a$ is a real number, defined by the "horizontal line" in your question so that the equation of the horizontal line is $x=a$. $\endgroup$ – Rory Daulton Sep 15 '17 at 0:23
  • $\begingroup$ @ArthurAlexKarapetov i do not like at all the comment wrote to the first answerer Jonathan similar to what you wrote under my answer..It seems that you do not know how math works according to these comments..Math is also reasoning not just ''throwing dry symbols in a blank parer" $\endgroup$ – Marios Gretsas Sep 15 '17 at 0:36
  • $\begingroup$ The other answerer provided a proof or at least a helpfull hint ,but just let you to fill in the details..he is not obligated to do your homework $\endgroup$ – Marios Gretsas Sep 15 '17 at 0:39

The number of times a horizontal line $y=y_0$ intersects the graph of a polynomial $p(x)$ is just the number of distinct solutions to the equation $p(x)=y_0$, or in other words the number of distinct roots of $p(x)-y_0=0$. So assume that $p(x)-y_0$ has $m$ distinct roots $x_1,x_2,\ldots,x_m$. Then you can divide (with zero remainder) $p(x)-y_0$ by $x-x_1$, $x-x_2$, $\ldots$, and $x-x_m$, which means that $$ p(x)-y_0=(x-x_1)(x-x_2)\cdots(x-x_m)q(x) $$ for some polynomial $q(x)$ with $\deg(q)=\deg(p-y_0)-m$. As $\deg(q)\geq0$ it follows $\deg(p-y_0)\geq m$, which implies $\deg(p)\geq m$.

Notice that it actually can happen that $\deg(p)>m$ (with $p(x)=x^3$ for instance).

  • $\begingroup$ But this explanation doesn't "prove" anything mathematically. It's just a compilation of assumptions. Could you please help with "proving it" as in writing a proof for the question? $\endgroup$ – Arthur Alex Karapetov Sep 15 '17 at 0:10
  • $\begingroup$ @ArthurAlexKarapetov I had actually misunderstood part of the question and that is why I edited my answer. Now it should be fine and clear. Also note that the only thing I am using is the fact that a polynomial $p(x)$ with coefficients in a field which has a root at $x_0$ can be divided without remainder by $x-x_0$. $\endgroup$ – Jonatan B. Bastos Sep 15 '17 at 1:13

Suppose that the degree of $P(x)=0$ is $d<m$

So from the fundamental theorem of algrebra $P(x)$ has $d$ roots so at most $d$ real roots.

Thus the horizontal line $y=0$ intersects the graph of $P(x)$ at most $d<m$ points which contradicts our hypothesis.

Can you generalize it for an arbitrary horizontal line $y=c$?

  • $\begingroup$ This doesn't "prove" anything mathematically. Could you please show how to prove what's beng asked? $\endgroup$ – Arthur Alex Karapetov Sep 15 '17 at 0:12
  • $\begingroup$ Mathematically??what do you think this proves?? I proved $\text{mathematically}$ by contradiction(if you are familair with it) that the degree is at least $m$ for the case when $y=0$..the points of this intersection are the zeros of the polynomia? $\endgroup$ – Marios Gretsas Sep 15 '17 at 0:15
  • $\begingroup$ Sorry, I'm not well versed in mathematics. It just seems to me this is not the answer my professor requires. $\endgroup$ – Arthur Alex Karapetov Sep 15 '17 at 0:16
  • $\begingroup$ What answer does your proffesor require then? $\endgroup$ – Marios Gretsas Sep 15 '17 at 0:19
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    $\begingroup$ @ArthurAlexKarapetov: This proof, as well as the other one, assume knowledge of The Fundamental Theorem of Algebra. If you know that theorem, your problem is easy. However, proving that theorem is most definitely not easy--see the link. Your professor almost certainly expects you to know and use the theorem. $\endgroup$ – Rory Daulton Sep 15 '17 at 0:26

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