Question

Write a quadratic function, f(x), in standard form with a leading coefficient of 1 for the given set of zeros: 3 and 4.

a) f(x) = (x - 3)(x - 4)
b) f(x) = x^2 - 7x + 12
c) f(x) = x^2 - 3x - 4
d) f(x) = (x - 2)(x - 7)

Answer 1

The quadratic function with a leading coefficient of 1 that has zeros at 3 and 4 is f(x) = x^2 - 7x + 12. This corresponds to option (b) in the provided choices. which corresponds to option (b).f(x) = x2 - 7x + 12,

Step-by-step explanation:

The student is asking for a quadratic function in standard form with zeros at 3 and 4. The standard form of a quadratic equation is ax2 + bx + c = 0. Given the zeros, we can use the factored form f(x) = (x - zero1)(x - zero2) to create the function.

To find the correct quadratic function, we substitute the zeros into the factored form to get f(x) = (x - 3)(x - 4). Expanding this we get:

f(x) = x2 - 4x - 3x + 12

Cleaning up the equation by combining like terms, we have:

f(x) = x2 - 7x + 12

So the correct quadratic function in standard form is f(x) = x2 - 7x + 12, which corresponds to option (b).