A polynomial function that does not intercept the x-axis has complex roots only
your example is wrong, since it intercepts the x-axis at x-1.
By definition, a graph which does not intercept the x-axis has no real roots -- that is, any roots are complex.
One exception might be y=3. It does not have any x-intercepts, but also has no complex roots.
A polynomial function that does not intercept the x-axis has complex roots only if all of its roots are complex numbers. In other words, there are no real roots.
One example of such a polynomial function is:
f(x) = (x - 1)(x - 2)(x - 3)
In this example, the polynomial has three roots: 1, 2, and 3. All of these roots are real numbers. Thus, this function does not have complex roots only.
To create a polynomial function with complex roots only, we can introduce the imaginary unit "i" into the polynomial.
For example, let's consider the polynomial:
g(x) = (x - 1)(x - i)(x + i)
In this example, the polynomial has three roots: 1, i, and -i. Both i and -i are complex numbers, as they involve the imaginary unit. Thus, this polynomial function does not intercept the x-axis and has complex roots only.
You are correct, and I apologize for the mistake in my previous example. The polynomial function (x - 1)(x - i)(x + i) does indeed intercept the x-axis at x = 1.
Thank you for pointing that out.
To provide a correct example, let's consider the simplest possible case of a polynomial that does not intercept the x-axis and has complex roots only:
f(x) = (x + i)(x - i)
In this example, the polynomial has two complex roots: i and -i. There are no real roots, and therefore f(x) does not intercept the x-axis.