How do you write the quadratic in vertex form given #y=5x^2+5x-3#?

Question

Julia Adams · Accepted Answer

To be in vertex form a quadratic needs to be expressed as
#y=color(red)(m)(x-color(green)(a))^2+color(orange)(b)#
(where #(a,b)# is the vertex of the quadratic), #y=5x^2+5x-3# #y = color(red)(5)(x^2+x)-3 "      extract the "color(red)( m)" term"# #y = color(red)(5)(color(blue)(x^2+x+(1/2)^2) -(1/2)^2)-3 "      "color(blue)("complete the square")# #y=color(red)5(x+1/2)^2 - 5/4 -3# #y=color(red)(5)(x- color(green)( (-1/2)))^2 + color(orange)(( -4 1/4))#

Hazel Anglin · Answer

undefined To write the quadratic in vertex form given $ y = 5x^2 + 5x - 3 $, follow these steps:

1. Complete the square for the quadratic expression.
2. Rewrite the quadratic expression in the form $ a(x - h)^2 + k $, where $ h $ and $ k $ are the coordinates of the vertex.

Starting with the given quadratic expression:

$$ y = 5x^2 + 5x - 3 $$

1. Complete the square for the quadratic expression:
$$ y = 5(x^2 + x) - 3 $$

To complete the square, add and subtract $\left(\frac{5}{2}\right)^2 = \frac{25}{4}$ inside the parentheses:
$$ y = 5(x^2 + x + \frac{25}{4} - \frac{25}{4}) - 3 $$

2. Rewrite the quadratic expression in the form $ a(x - h)^2 + k $:
$$ y = 5\left(x^2 + x + \frac{25}{4}\right) - 5\left(\frac{25}{4}\right) - 3 $$
$$ y = 5\left(x + \frac{1}{2}\right)^2 - \frac{125}{4} - 3 $$
$$ y = 5\left(x + \frac{1}{2}\right)^2 - \frac{137}{4} $$

Therefore, the quadratic equation in vertex form is $ y = 5\left(x + \frac{1}{2}\right)^2 - \frac{137}{4} $.