The function given is f(x) = x² – 6x + 3. To rewrite this function in vertex form, we need to complete the square.
The standard vertex form of a quadratic function is:
f(x) = a(x – h)² + k
Where (h, k) is the vertex of the parabola. Let’s start with the function:
f(x) = x² – 6x + 3
First, focus on the quadratic and linear terms: x² – 6x. To complete the square, take the coefficient of x (which is -6), divide it by 2, and square it:
- (-6 / 2)² = (-3)² = 9
Next, add and subtract this value (9) inside the function:
f(x) = x² – 6x + 9 – 9 + 3
Now we can group x² – 6x + 9, which is a perfect square:
f(x) = (x – 3)² – 6
So, now we have f(x) in vertex form as:
f(x) = (x – 3)² – 6
The vertex of this parabola is at the point (3, -6). This means the parabola opens upwards (since the coefficient of the squared term, a = 1, is positive) and has its vertex at the point (3, -6).
Thus, the vertex form equivalent to the given function f(x) = x² – 6x + 3 is:
f(x) = (x – 3)² – 6