An equation for a quartic function with zeros 4, 5, and 6 that passes through the point (7, 18) is

One example of this can be:

P(x) = (3/2)*(x - 4)*(x - 5)*(x - 5)*(x - 6)

Step-by-step explanation:

A quartic equation is a polynomial of degree 4.

Now, remember that for a polynomial of degree n, with leading coefficient A and zeros {x₁, x₂, ..., xₙ}

The polinomial can be written as:

p(x) = A*(x - x₁)*(x - x₂)*...*(x - xₙ)

In this case we know that we have the zeros 4, 5 and 6.

Notice that this is a polynomial of degree 4 but we have 3 zeros, so one of them may be a double one, i will assume that is the 5.

And we have a leading coefficient that we do not know, let's call it A

Then we can write our polynomial as:

P(x) = A*(x - 4)*(x - 5)*(x - 5)*(x - 6)

Now we know that the polynomial passes through the point (7, 18), then:

P(7) = 18 = A*(7 - 4)*(7 - 5)*(7 - 5)*(7 - 6)

With this equation, we can find the value of A.

18 = A*(7 - 4)*(7 - 5)*(7 - 5)*(7 - 6)

18 = A*12

18/12 = A

(3/2) = A

Then our equation can be:

P(x) = (3/2)*(x - 4)*(x - 5)*(x - 5)*(x - 6)