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

1 Answer
May 16, 2016

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.