To find a cubic function that has real zeros at x = 3 and x = 7, we start by recognizing that a cubic function can have up to three real zeros. Since we only have two specified zeros, we can introduce a third zero, which can be any real number, including possibly being a repeated zero.
Given the zeros we have, we can express the function in factored form:
f(x) = k(x - 3)(x - 7)(x - r)
Here, k is a constant that can stretch or compress the graph, and r is the third zero. For simplicity, we can choose r = 0, leading to the function:
f(x) = k(x - 3)(x - 7)x
If we set k = 1, we have:
f(x) = (x - 3)(x - 7)x
Expanding this gives:
f(x) = x(x - 3)(x - 7) = x(x^2 - 10x + 21) = x^3 - 10x^2 + 21x
Thus, one possible cubic function with real zeros at x = 3 and x = 7 is:
f(x) = x^3 - 10x^2 + 21x
This function meets the criteria and has the desired zeros.