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

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.

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