To rewrite the quadratic function f(x) = 3x² + 6x + 9 in vertex form, we start by identifying the standard form of a quadratic, which is given by:
f(x) = a(x – h)² + k
where (h, k) is the vertex of the parabola. The first step in this process is to factor out the coefficient of x² from the first two terms.
1. Factor out 3 from the first two terms:
f(x) = 3(x² + 2x) + 9
2. Next, we need to complete the square for the expression inside the parentheses. To do this, we take the coefficient of x (which is 2), divide it by 2 (giving us 1), and then square it (resulting in 1).
3. Add and subtract that square inside the parentheses:
f(x) = 3(x² + 2x + 1 – 1) + 9
4. This simplifies to:
f(x) = 3((x + 1)² – 1) + 9
5. Distributing the 3 gives us:
f(x) = 3(x + 1)² – 3 + 9
6. Combine the constants:
f(x) = 3(x + 1)² + 6
Now we have the vertex form of the quadratic function:
f(x) = 3(x + 1)² + 6
In this form, the vertex of the parabola is located at the point (-1, 6).