# Write the quadratic function f(x) = x2 + 8x + 3 in vertex form? A) f(x) = (x - 4)2 - 13 B) f(x) = (x - 4)2 + 3 C) f(x) = (x + 4)2 + 3 D) f(x) = (x + 4)2 - 13

Mar 27, 2016

$\text{D} : f \left(x\right) = {\left(x + 4\right)}^{2} - 13$

#### Explanation:

Given the following function, you are asked to convert it to vertex form:

$f \left(x\right) = {x}^{2} + 8 x + 3$

The given possible solutions are:

"A") f(x)=(x-4)^2-13
"B") f(x)=(x-4)^2+3
"C") f(x)=(x+4)^2+3
"D") f(x)=(x+4)^2-13

Converting to Vertex Form
$1$. Start by placing brackets around the first two terms.

$f \left(x\right) = {x}^{2} + 8 x + 3$

$f \left(x\right) = \left({x}^{2} + 8 x\right) + 3$

$2$. In order to make the bracketed terms a perfect square trinomial, we must add a "$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{c}$" term as in $a {x}^{2} + b x + \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{c}$. Since $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{c}$, in a perfect square trinomial is denoted by the formula $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{c} = {\left(\frac{\textcolor{b l u e}{b}}{2}\right)}^{2}$, take the value of $\textcolor{b l u e}{b}$ to find the value of $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{c}$.

$f \left(x\right) = \left({x}^{2} + \textcolor{b l u e}{8} x + {\left(\frac{\textcolor{b l u e}{8}}{2}\right)}^{2}\right) + 3$

$3$. However, adding ${\left(\frac{8}{2}\right)}^{2}$ would change the value of the equation. Thus, subtract ${\left(\frac{8}{2}\right)}^{2}$ from the ${\left(\frac{8}{2}\right)}^{2}$ you just added.

$f \left(x\right) = \left({x}^{2} + 8 x + {\left(\frac{8}{2}\right)}^{2} - {\left(\frac{8}{2}\right)}^{2}\right) + 3$

$4$. Multiply $\left(- {\left(\frac{8}{2}\right)}^{2}\right)$ by the $\textcolor{v i o \le t}{a}$ term as in $\textcolor{v i o \le t}{a} {x}^{2} + b x + c$ to bring it outside the brackets.

$f \left(x\right) = \left(\textcolor{v i o \le t}{1} {x}^{2} + 8 x + {\left(\frac{8}{2}\right)}^{2}\right) + 3 - \left({\left(\frac{8}{2}\right)}^{2} \times \textcolor{v i o \le t}{1}\right)$

$5$. Simplify.

$f \left(x\right) = \left({x}^{2} + 8 x + 16\right) + 3 - 16$

$f \left(x\right) = \left({x}^{2} + 8 x + 16\right) - 13$

$6$. Lastly, factor the perfect square trinomial.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} f \left(x\right) = {\left(x + 4\right)}^{2} - 13 \textcolor{w h i t e}{\frac{a}{a}} |}}}$