Answer:
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)
