In the xy plane (x, y) is on the x axis and
y =x^4 - 15x^2 +14, then (x, y) could be

One  answer is (x, y) = (1, 0).

See below.

Step-by-step explanation:

If (x,y) is on the x axis then y = 0.

When y = 0.

x^4 - 15x^2 + 14 = 0

Let x^2 = z, then:

z^2 - 15z + 14 = 0

(z - 1)(z - 14) = 0

So z = 1, 14.

But z = x^2,

so x could be +/- 1 or +/-√14

and (x, y) could be (1, 0), (-1, 0), (√14, 0) and (-√14, 0).

These are the zeroes of the function.