What is the vertex form of y= 4x^2 + 17x + 4 ?

Jan 10, 2016

$y = 4 {\left(x + \frac{17}{8}\right)}^{2} - 140.5$

Explanation:

First, find x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{17}{8}$
Next, find y-coordinate of vertex
y(-17/8) = 4(289/64) - 17(17/8) + 4 = 1156/64 - 289/8 + 4 = = -1156/8 + 32/8 = - 1124/8 = -140.5
Vertex form:
$y = 4 {\left(x + \frac{17}{8}\right)}^{2} - 140.5$