# What is the vertex form of y=6x^2+13x+3 ?

May 16, 2016

The general formula for vertex form is
$y = a {\left(x - \left(- \frac{b}{2 a}\right)\right)}^{2} + c - {b}^{2} / \left\{4 a\right\}$

y=6(x-(-13/{2*6}))^2+3 -13^2/{4*6})
$y = 6 {\left(x - \left(- \frac{13}{12}\right)\right)}^{2} + \left(- \frac{97}{24}\right)$

$y = 6 {\left(x - \left(- 1.08\right)\right)}^{2} + \left(- 4.04\right)$

You can also find the answer by completing the square, the general formula is found by completing the square in using $a {x}^{2} + b x + c$. (see below)

#### Explanation:

The vertex form is given by
$y = a {\left(x - {x}_{v e r t e x}\right)}^{2} + {y}_{v e r t e x}$,
where $a$ is the "stretch" factor on the parabola and the coordinates of the vertex is $\left({x}_{v e r t e x} , {y}_{v e r t e x}\right)$

This form highlights the transformations that the function $y = {x}^{2}$underwent to build that particular parabola, shifting to the right by ${x}_{v e r t e x}$, up by ${y}_{v e r t e x}$ and stretched /flipped by $a$.

The vertex form is also form in which a quadratic function can be directly solved algebraically (if it has a solution). So getting a quadratic function into vertex form from standard form, called completing the square, is the first step to solving the equation.

The key to completing the square is building a perfect square in ANY quadratic expression. A perfect square is of the form
$y = {\left(x + p\right)}^{2} = {x}^{2} + 2 \cdot p + {p}^{2}$

Examples
${x}^{2} + 24 x + 144$ is a perfect square, equal to ${\left(x + 12\right)}^{2}$
${x}^{2} - 12 x + 36$ is a perfect square, equal to ${\left(x - 6\right)}^{2}$
$4 {x}^{2} + 36 x + 81$ is a perfect square, equal to ${\left(2 x + 9\right)}^{2}$

COMPLETING THE SQUARE
$y = 6 {x}^{2} + 13 x + 3$
factor out the 6
$y = 6 \left({x}^{2} + \frac{13}{6} x\right) + 3$
Multiply and divide the linear term by 2
$y = 6 \left({x}^{2} + 2 \cdot \left(\frac{13}{12}\right) x\right) + 3$

This lets us see what our $p$ has to be, HERE $p = \left(\frac{13}{12}\right)$.
To build our perfect square we need the ${p}^{2}$ term, ${13}^{2} / {12}^{2}$
we add this to our expression, but to avoid changing the value of anything we must subtract it too, this creates an extra term, $- {13}^{2} / {12}^{2}$.
$y = 6 \left({x}^{2} + 2 \cdot \left(\frac{13}{12}\right) x + \frac{{13}^{2}}{{12}^{2}} - \frac{{13}^{2}}{{12}^{2}}\right) + 3$
We gather up our perfect square
$y = 6 \left(\left({x}^{2} + 2 \cdot \left(\frac{13}{12}\right) x + \frac{{13}^{2}}{{12}^{2}}\right) - \frac{{13}^{2}}{{12}^{2}}\right) + 3$
and replace it with ${\left(x + p\right)}^{2}$, HERE ${\left(x + \frac{13}{12}\right)}^{2}$
$y = 6 \left({\left(x + \frac{13}{12}\right)}^{2} - \frac{{13}^{2}}{{12}^{2}}\right) + 3$

We multiple out our extra to to get it outside the brackets.
$y = 6 {\left(x + \frac{13}{12}\right)}^{2} - 6 \frac{{13}^{2}}{{12}^{2}} + 3$
Play with some fractions to neaten
$y = 6 {\left(x + \frac{13}{12}\right)}^{2} - \frac{6 \cdot {13}^{2}}{12 \cdot 12} + \frac{3 \cdot 12 \cdot 12}{12 \cdot 12}$
$y = 6 {\left(x + \frac{13}{12}\right)}^{2} + \frac{3 \cdot 12 \cdot 12 - 6 \cdot 13 \cdot 13}{12 \cdot 12}$
And we have
$y = 6 {\left(x + \frac{13}{12}\right)}^{2} - \frac{97}{24}$.

If we want to in the identical form as above
$y = a {\left(x - {x}_{v e r t e x}\right)}^{2} + {y}_{v e r t e x}$, we gather up the signs as so
$y = 6 {\left(x - \left(- \frac{13}{12}\right)\right)}^{2} + \left(- \frac{582}{144}\right)$.

The general formula used above is from doing the above with $a {x}^{2} + b x + c$ and is the first step to proving the quadratic formula.