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

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

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Answer 1 · 2016-05-16T20:15:20

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

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

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

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

Explanation:

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

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

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
$y = {(x + p)}^{2} = x^{2} + 2 \cdot p + p^{2}$ $y=(x+p)^2=x^2+2*p+p^2$

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

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

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

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

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

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

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

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What is the vertex form of y=6x^2+13x+3 y=6x2+13x+3y=6x^2+13x+3 ?

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