Question

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

Answer 1

The general formula for vertex form is
`y=a(x-(-b/{2a}))^2+ c-b^2/{4a}`

`y=6(x-(-13/{2*6}))^2+3 -13^2/{4*6})`
`y=6(x-(-13/12))^2+(-97/24)`

`y=6(x-(-1.08))^2+(-4.04)`

You can also find the answer by completing the square, the general formula is found by completing the square in using `ax^2+bx+c`. (see below)

Explanation:

The vertex form is given by
`y=a(x-x_{vertex})^2 + y_{vertex}`,
where `a` is the "stretch" factor on the parabola and the coordinates of the vertex is `(x_{vertex},y_{vertex})`

This form highlights the transformations that the function `y=x^2`underwent to build that particular parabola, shifting to the right by `x_{vertex}`, up by `y_{vertex}` 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=(x+p)^2=x^2+2*p+p^2`

Examples
`x^2 + 24x +144` is a perfect square, equal to `(x+12)^2`
`x^2 - 12x +36` is a perfect square, equal to `(x-6)^2`
`4x^2 + 36x +81` is a perfect square, equal to `(2x+9)^2`

COMPLETING THE SQUARE
You start with
`y=6x^2+13x+3`
factor out the 6
`y=6(x^2+13/6x)+3`
Multiply and divide the linear term by 2
`y=6(x^2+2*(13/12)x)+3`

This lets us see what our `p` has to be, HERE `p=(13/12)`.
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(x^2+2*(13/12)x+{13^2}/{12^2}-{13^2}/{12^2})+3`
We gather up our perfect square
`y=6((x^2+2*(13/12)x+{13^2}/{12^2})-{13^2}/{12^2})+3`
and replace it with `(x+p)^2`, HERE `(x+13/12)^2`
`y=6((x+13/12)^2-{13^2}/{12^2})+3`

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

If we want to in the identical form as above
`y=a(x-x_{vertex})^2 + y_{vertex}`, we gather up the signs as so
`y=6(x-(-13/12))^2+(-582/144)`.

The general formula used above is from doing the above with `ax^2+bx+c` and is the first step to proving the quadratic formula.