What is the vertex form of y= 3x^2+29x-44 y=3x2+29x−44?

1 Answer
Isini M.
Jun 8, 2017

y=3(x+29/6)^2-1369/12y=3(x+296)2−136912

Explanation:

Method 1 - Completing the Square
To write a function in vertex form (y=a(x-h)^2+ky=a(x−h)2+k), you must complete the square.
y=3x^2+29x-44y=3x2+29x−44

  1. Make sure you factor out any constant in front of the x^2x2 term, i.e. factor out the aa in y=ax^2+bx+cy=ax2+bx+c.
    y=3(x^2+29/3x)-44y=3(x2+293x)−44

  2. Find the h^2h2 term (in y=a(x-h)^2+ky=a(x−h)2+k) that will complete the perfect square of the expression x^2+29/3xx2+293x by dividing 29/3293 by 22 and squaring this.
    y=3[(x^2+29/3x+(29/6)^2)-(29/6)^2]-44y=3[(x2+293x+(296)2)−(296)2]−44
    Remember, you cannot add something without adding it to both sides, that is why you can see (29/6)^2(296)2 subtracted.

  3. Factorise the perfect square:
    y=3[(x+29/6)^2-(29/6)^2]-44y=3[(x+296)2−(296)2]−44

  4. Expand brackets:
    y=3(x+29/6)^2-3×841/36-44y=3(x+296)2−3×84136−44

  5. Simplify:
    y=3(x+29/6)^2-841/12-44y=3(x+296)2−84112−44
    y=3(x+29/6)^2-1369/12y=3(x+296)2−136912

Method 2 - Using General Formula
y=a(x-h)^2+ky=a(x−h)2+k
h=-b/(2a)h=−b2a
k=c-b^2/(4a)k=c−b24a
From your question, a=3, b=29, c=-44a=3,b=29,c=−44
Therefore, h=-29/(2×3)h=−292×3
h=-29/6h=−296
k=-44-29^2/(4×3)k=−44−2924×3
k=-1369/12k=−136912
Substituting aa, hh and kk values into general vertex form equation:
y=3(x-(-29/6))^2-1369/12y=3(x−(−296))2−136912
y=3(x+29/6)^2-1369/12y=3(x+296)2−136912

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