Vrftjkry

I know that a quartic function has 4 x-intercepts. But, can it have 2 x-intercepts? If so, what would a quartic function that has 2 x-intercepts look like?

Take $f(x)=(x^2-1)^2$ or $(x^2-1)(x^2+1)$.

A polynomial of degree $n$ has at most $n$ solutions or "zeros" (values at which the polynomial equal $0$) or $x$- intercepts. So you want a quartic with exactly two zeros.

You can have:

$Q(x) = (x -m)(x -n)(ax^2 + bx + c)$ where $ax^2 + bx + c$ has no real roots. (i.e. if $b^2 - 4ac < 0$)

So for example $(x-1)(x + 1)(x^2 + x + 1)$. That will have roots at $x =1;x =-1$

If you graph this this will look like a W where $x$ axis passes above (or below) the middle hump.

Or you can multiple roots of the from $(x-m)^k$.

for a quartic you can have $Q(x) = a(x-m)^2(x-n)^2$ which will have double root at $m$ or $n$.

If you graph thus it will look like a W where the $x$ axis goes through the bottommost points of the W.

Or you can have $Q(x) = a(x-m)^3(x-n)$ which will have a triple root at $m$ and a single root at $n$.

This will have ... Well, at $m$ the graph comes in and flattens, but then goes to the other side. There will be a hump at that point and the graph will come back through.

But you can't have $Q(x) = a(x-m)^2(x-n)P(x)$, a double and a triple root where $P(x)$ has no root because $P(x) = x +c$ and that always has a root.

Yes, a Quartic(aka biquadratic) polynomial's graph can have two $x$ intercepts.
We use the fact that a polynomial of degree $n$ can have at most $n$ roots(aka "zeroes"), or, $n$ number of $x$ intercepts for the graph. Thus, a biquadratic polynomial can have at most four $x$ intercepts, depending on the number of real roots. If all four of its roots are real, then the graph will have four $x$ intercepts;if it has three real roots , then the graph will have three $x$ intercepts , and so on.
Thus, your condition is possible if the biquadratic has 2 real roots and two imaginary roots (i.e. involving $i$ (iota) which is equal to the square root of $-1$).
For example, the graph of $f(x)=(x^2-4)(x^2+4)$ will have only two $x$ intercepts. So we can say, for $f(x)=(x^2-k)(x^2+k)$ (where $k$ is a positive number), the graph will have only two $x$ intercepts.