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

Jun 8, 2017

$y = 3 {\left(x + \frac{29}{6}\right)}^{2} - \frac{1369}{12}$

#### Explanation:

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

1. Make sure you factor out any constant in front of the ${x}^{2}$ term, i.e. factor out the $a$ in $y = a {x}^{2} + b x + c$.
$y = 3 \left({x}^{2} + \frac{29}{3} x\right) - 44$

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

3. Factorise the perfect square:
$y = 3 \left[{\left(x + \frac{29}{6}\right)}^{2} - {\left(\frac{29}{6}\right)}^{2}\right] - 44$

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

5. Simplify:
$y = 3 {\left(x + \frac{29}{6}\right)}^{2} - \frac{841}{12} - 44$
$y = 3 {\left(x + \frac{29}{6}\right)}^{2} - \frac{1369}{12}$

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