What is the vertex form of #y= 3x^2+29x-44 #?

1 Answer
Jun 8, 2017

#y=3(x+29/6)^2-1369/12#

Explanation:

Method 1 - Completing the Square
To write a function in vertex form (#y=a(x-h)^2+k#), you must complete the square.
#y=3x^2+29x-44#

  1. Make sure you factor out any constant in front of the #x^2# term, i.e. factor out the #a# in #y=ax^2+bx+c#.
    #y=3(x^2+29/3x)-44#

  2. Find the #h^2# term (in #y=a(x-h)^2+k#) that will complete the perfect square of the expression #x^2+29/3x# by dividing #29/3# by #2# and squaring this.
    #y=3[(x^2+29/3x+(29/6)^2)-(29/6)^2]-44#
    Remember, you cannot add something without adding it to both sides, that is why you can see #(29/6)^2# subtracted.

  3. Factorise the perfect square:
    #y=3[(x+29/6)^2-(29/6)^2]-44#

  4. Expand brackets:
    #y=3(x+29/6)^2-3×841/36-44#

  5. Simplify:
    #y=3(x+29/6)^2-841/12-44#
    #y=3(x+29/6)^2-1369/12#

Method 2 - Using General Formula
#y=a(x-h)^2+k#
#h=-b/(2a)#
#k=c-b^2/(4a)#
From your question, #a=3, b=29, c=-44#
Therefore, #h=-29/(2×3)#
#h=-29/6#
#k=-44-29^2/(4×3)#
#k=-1369/12#
Substituting #a#, #h# and #k# values into general vertex form equation:
#y=3(x-(-29/6))^2-1369/12#
#y=3(x+29/6)^2-1369/12#