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^2underwent 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.