How to Construct Polynomials with known zeros

Question

How can I construct a polynomial with the x-intercepts at 0,2,5 where the graph of f(x) crosses the x axis at 0, but does not cross it at 2 or 5?

Answer 1

I was about to say what you asked was impossible because "x-intercept" MEANS crosses the x-axis.

Then I read Fakemistake's answer and realize "x-intercept" means touches the x-axis.

If it touches or crosses the x-axis at $x = a$ than $a$ is a root and $x-a$ is a factor. If $a$ is an odd-power root, the graph crosses but if $a$ is an even power, it touches but does not cross.

So $P(x) = x(x-2)^2(x-5)^2$ would be the simplest polynomial that does what you want. But any $S(x)= c*[\prod (x-a_i)]^{k_i}] x^m(x-2)^{2n}(x-5)^{2j}$ where none of the $a_i = 0,2$ or $5$, will also work.

However $P(x)=x(x-2)^2(x-5)^2$ with a power of $5$ is the simplest solution as it has the least power possible and a leading coefficient of $1$.

Answer 2

The graph of $f(x)=(x-0)(x-2)^2(x-5)$ crosses the x-axis in $x=0$ and $x=5$ but not in $x=2$.

And more general: Consider the graph of $(x-0)^{k_0}(x-2)^{k_2}(x-5)^{k_5}$. The graph doesn't cross the $x$-axis in $x=0$ if $k_0$ is even. The same is true for the pairs $(x=2,k_2)$ and $(x=5,k_5)$. If $k_i$ is odd, then the graph crosses the x-axis in this root.

Answer 3

$p(x)=x(x-2)^2(x-5)^2 $ is such a polynomial.

Note that when the exponent is even the graph does not cross the x axis.