$$f(x)= x^2+4(x-3) $$

Which of the following is an equivalent form of the expression above that displays the vertex of the function as constants or coefficients in the expression?

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There are three primary forms of the quadratic function:

$$ f(x)=x^2+4(x-3) $$

We can expand the given equation to obtain the standard form \(f(x)=ax^2+bx+c\).

$$f(x)=x^2+4x-12 \tag*{\tiny standard form}$$

The standard form displays the \(y\)-intercept.

$$f(x)=x^2+4x+\colorbox{aqua}{$-12$} $$

Factoring the equation results in the factored form \( f(x)=a(x-b)(x-c) \).

$$ f(x)=(x+6)(x-4) \tag*{\tiny factored form} $$

The factored form displays the \(x\)-intercepts.

$$ f(x)=(x-\colorbox{aqua}{$-6$})(x-\colorbox{aqua}{4}) $$

Using the standard form, we can complete the square to obtain the vertex form \(f(x)=a(x-h)^2+k \).

$$f(x)=x^2+4x-12 $$ $$ f(x)= \underbrace{x^2 + 4x \colorbox{yellow}{$+4$}}_\text{perfect square trinomial} -12 \colorbox{yellow}{$-4$} $$ $$ f(x)=(x+2)^2-16 \tag*{\tiny vertex form}$$

The vertex form displays the vertex.

$$ f(x)=(x-\colorbox{aqua}{$-2$})^2+\colorbox{aqua}{$-16$} $$