Question

write $f(x) = 2x^2 - 44x + 185$ in vertex form.
to write $f(x) = 2x^2 - 44x + 185$, factor out \\(\square\\) from the first two terms.
next, form a perfect square trinomial keeping the value of the function equivalent:
$f(x) = 2(x^2 - 22x + 121) + 185 - 242$
the function written in vertex form is $f(x) = \square (x - \square )^2 + \square$.

Explanation:

Step1: Factor out coefficient of \(x^2\)

The first two terms are \(2x^2 - 44x\). The coefficient of \(x^2\) is 2, so we factor out 2: \(2(x^2 - 22x)\).

Step2: Complete the square inside the parentheses

For the quadratic \(x^2 - 22x\), to complete the square, we take half of -22, which is -11, and square it: \((-11)^2 = 121\). So we add and subtract 121 inside the parentheses, but since there's a factor of 2 outside, we actually add \(2\times121\) and subtract \(2\times121\) to keep the function equivalent. We already have \(2(x^2 - 22x + 121) + 185 - 242\) (because \(2\times121 = 242\)).

Step3: Write in vertex form

The vertex form of a quadratic function is \(f(x) = a(x - h)^2 + k\), where \((h, k)\) is the vertex. From \(2(x^2 - 22x + 121) + 185 - 242\), we can write \(x^2 - 22x + 121\) as \((x - 11)^2\). Then simplify the constant term: \(185 - 242 = -57\). So the function becomes \(f(x) = 2(x - 11)^2 - 57\).

Answer:

First blank: 2; Vertex form: \(f(x) = \boldsymbol{2}(x - \boldsymbol{11})^2 + \boldsymbol{(-57)}\) (or \(f(x) = 2(x - 11)^2 - 57\))