Question

Why is f(t)=(3t+3)^2 not the vertex form of f(x)=9x^2+2x+1?

a. The variable x has a coefficient.
b. The expression is not the product of two binomials.
c. Some of the terms are tions instead of integers.
d. The expression has a constant outside of the squared term.

Answer 1

The expression f(t) = (3t+3)^2 is not the vertex form of f(x) = 9x^2 + 6x + 1 because the expanded form 9t^2 + 18t + 9 does not equal the original quadratic equation 9x^2 + 6x + 1. The coefficients of the linear term and the constants are different.

Step-by-step explanation:

The expression f(t) = (3t+3)^2 is not the vertex form of f(x) = 9x^2 + 6x + 1 because while they are related, they are not mathematically equivalent. The vertex form of a quadratic equation is typically given as f(x) = a(x-h)^2 + k, where (h, k) is the vertex of the parabola.

In the case of f(x) = 9x^2 + 6x + 1, completing the square would give us the equivalent expression where the coefficient of x is factored out, and a constant term is added to achieve the perfect square trinomial needed for vertex form, which would look something like f(x) = a(x-h)^2 + k. For the given quadratic, the vertex form would be f(x) = 9(x+1/3)^2, indicating that the vertex is at (-1/3, 0).

The given expression f(t) = (3t+3)^2 would expand to 9t^2 + 18t + 9, which is not the same as 9x^2 + 6x + 1. The coefficient of the linear term and the constant term both differ, hence, (3t+3)^2 is not the vertex form of f(x) = 9x^2 + 6x + 1.