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Rewrite the quadratic function f(x) = 6x² - 6x + 1 in vertex form.

User Korvo
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Final answer:

To rewrite the quadratic f(x) = 6x² - 6x + 1 in vertex form, we complete the square to get f(x) = 6(x - 1/2)² - 1/2, where the vertex of the parabola is at (1/2, -1/2).

Step-by-step explanation:

To rewrite the quadratic function f(x) = 6x² - 6x + 1 in vertex form, we need to complete the square. The vertex form of a quadratic function is f(x) = a(x - h)²9 + k, where (h, k) is the vertex of the parabola. First, we factor out the coefficient of the x² term from the first two terms:

f(x) = 6(x² - x) + 1

Next, we find the value that completes the square. The value to add and subtract inside the parenthesis is (½²), here it's (½*-1)² = 1/4. We add and subtract this value inside the parentheses:

f(x) = 6(x² - x + 1/4 - 1/4) + 1

Combine the +1/4 with the x terms and take the -1/4 outside the parenthesis multiplied by 6:

f(x) = 6((x - 1/2)² - 1/4) + 1

Now, distribute the 6:

f(x) = 6(x - 1/2)² - 6/4 + 1

Simplify the constants:

f(x) = 6(x - 1/2)² - 3/2 + 2/2

f(x) = 6(x - 1/2)² - 1/2

So, the vertex form of f(x) = 6x² - 6x + 1 is f(x) = 6(x - 1/2)² - 1/2 where (h, k) = (1/2, -1/2)

User Ofer Zelig
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