What is the vertex form of #y= 3x^2+29x-44 #?
1 Answer
Explanation:
Method 1 - Completing the Square
To write a function in vertex form (
-
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# -
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. -
Factorise the perfect square:
#y=3[(x+29/6)^2-(29/6)^2]-44# -
Expand brackets:
#y=3(x+29/6)^2-3×841/36-44# -
Simplify:
#y=3(x+29/6)^2-841/12-44#
#y=3(x+29/6)^2-1369/12#
Method 2 - Using General Formula
From your question,
Therefore,
Substituting