# How do you rewrite the equation in vertex form: 4x^2+16-9?

May 4, 2015

Version 1: Where I assume an $x$ was accidentally missed on the $+ 16$ term:
$y = 4 {x}^{2} + 16 x - 9$

Vertex form of a quadratic is
$y = m {\left(x - a\right)}^{2} + b$
where the vertex of the parabola is at $\left(a , b\right)$

$y = 4 {x}^{2} + 16 x - 9$

$= 4 \left({x}^{2} + 4 x\right) - 9 \text{ extracting the "m" factor}$

$= 4 \left({x}^{2} + 4 x + {2}^{2}\right) - 16 - 9 \text{ [completing the square](http://socratic.org/algebra/quadratic-equations-and-functions/completing-the-square-1)}$

$= 4 \left(x + 2\right) - 25 \text{ simplifying}$

$= 4 \left(x - \left(- 2\right)\right) + \left(- 25\right) \text{ into vertex form}$

(The vertex is at $\left(x , y\right) = \left(- 2 , - 25\right)$)

Version 2: Where the question was entered correctly (except for the missing $\left(y =\right)$ which is needed to make it an equation
$y = 4 {x}^{2} + 16 - 9$
which is equivalent to $y = 4 {x}^{2} + 7$

This can be rearranged as
$y = 4 {\left(x - 0\right)}^{2} + 7$
for a parabola with a vertex at $\left(x , y\right) = \left(0 , 7\right)$