# What is the vertex form of y=4x^2+5x+2 ?

Mar 22, 2016

$y = 4 {\left(x + \frac{5}{8}\right)}^{2} + \frac{7}{16}$

#### Explanation:

The standard form of the quadratic function is :$y = a {x}^{2} + b x + c$

The function: $y = 4 {x}^{2} + 5 x + 2 \text{ is in this form }$

with a = 4 , b = 5 and c = 2
>$\text{--------------------------------------------------}$
The vertex form of the quadratic function is

$y = a {\left(x - h\right)}^{2} + k \text{ (h,k) are the coords of vertex }$

x-coord of vertex (h) $= - \frac{b}{2 a} = - \frac{5}{2 \times 4} = - \frac{5}{8}$
now substitute $x = - \frac{5}{8} \text{ into } y = 4 {x}^{2} + 5 x + 2$
y-coord of vertex (k) = $4 {\left(- \frac{5}{8}\right)}^{2} + 5 \left(- \frac{5}{8}\right) + 2$
$= 4 \left(\frac{25}{64}\right) - \frac{25}{8} + 2 = \frac{7}{16}$
hence vertex has coordinates $\left(- \frac{5}{8} , \frac{7}{16}\right)$
> $\text{------------------------------------------------}$
so a = 4 and (h , k ) $= \left(- \frac{5}{8} , \frac{7}{16}\right)$

$\Rightarrow \text{ vertex form is } y = 4 {\left(x + \frac{5}{8}\right)}^{2} + \frac{7}{16}$