Question

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

Answer 1

`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`