Use the rational zero theorem, which of the following could not be a rational zero of: 6x^5+ax^3=bx+12? 1/2 -3/4 -2

x = -3/4 can not be a rational zero of the polynomial.

Step-by-step explanation:

We have the polynomial:

6x^5 + ax^3 -bx -12 = 0.

The theorem says that:

If P(x) is a polynomial with integer coefficients, and p/q is a zero of P(x) then p is a factor of the constant term (in this case the constant term is -12) and q is a factor of the leading coefficient (in this case the leading coefficient is 6.).

The factors of -12  (different than itself) are (independent of the sign).

1, 2, 3, 4 and 6.

So p can be: 1, -1, 2, -2, 3, -3, 4, -4, 6, -6.

The factors of 6 are:

1, 2 and 3, so q can be 1, -1, 2, -2, 3, -3.

Then the option that can not be a zero of the polynomial is

x = -3/4

because the number in the denominator must be a factor of the leading coefficient, and 4 is not a factor of six.

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: