What is the vertex form of #y=6x^2-13x-5 #?

1 Answer
Mar 2, 2016

#y = 6(x - 13/12)^2 - 289/24 #

Explanation:

The standard form of the quadratic function is #ax^2+bx+c #

the function here #y = 6x^2-13x-5 " is in this form " #

by comparison , a = 6 , b = -13 and c = -5

The vertex form is : # y=a(x-h)^2 + k #

where (h,k) are the coords of the vertex.

the x-coord of the vertex (h)# = (-b)/(2a) = -(-13)/12 = 13/12#

and y-coord (k) #= 6(13/12)^2 -13(13/12) - 5 = -289/24 #

here #(h,k) = (13/12 , -289/24 ) and a = 6 #

#rArr y = 6(x-13/12)^2 - 289/24 " is the equation " #