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