What is the vertex form of $7 y = 3 x^{2} - 2 x + 12$ $7y=3x^2 − 2x + 12$ ?

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What is the vertex form of $7 y = 3 x^{2} - 2 x + 12$ $7y=3x^2 − 2x + 12$ ?

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Answer 1 · 2016-10-07T20:55:08

We will have to complete the square for this quadratic which will put the equation in vertex form.

First lets solve for the y variable by dividing both sides by 7

$\frac{7 y}{7} = \frac{3}{7} x^{2} - \frac{2}{7} x + \frac{12}{7}$ $(cancel7y)/cancel7=3/7x^2-2/7x+12/7$

Set the equation equal to zero.

$0 = \frac{3}{7} x^{2} - \frac{2}{7} x + \frac{12}{7}$ $0=3/7x^2-2/7x+12/7$

Subtract $\frac{12}{7}$ $12/7$ from both sides

$0 - \frac{12}{7} = \frac{3}{7} x^{2} - \frac{2}{7} x + \frac{12}{7} - \frac{12}{7}$ $0color(red)(-12/7)=3/7x^2-2/7x+12/7color(red)(-12/7)$

Simplify

$- \frac{12}{7} = \frac{3}{7} x^{2} - \frac{2}{7} x$ $color(red)(-12/7)=3/7x^2-2/7x$

Factor out $\frac{3}{7}$ $3/7$

$- \frac{12}{7} = \frac{3}{7} (x^{2} - \frac{2}{7} (\frac{7}{3}) x)$ $-12/7=3/7(x^2-2/cancel7(cancel7/3)x)$

Simplify

$- \frac{12}{7} = \frac{3}{7} (x^{2} - \frac{2}{3} x)$ $-12/7=3/7(x^2-2/3x)$

Take the coefficient of x and divide it by 2 and then square it

${(\frac{- \frac{2}{3}}{2})}^{2} = {(- \frac{2}{3} \cdot \frac{1}{2})}^{2} = {(- \frac{2}{6})}^{2} = {(- \frac{1}{3})}^{2} = \frac{1}{9}$ $((-2/3)/2)^2=(-2/3*1/2)^2=(-2/6)^2=(-1/3)^2=1/9$

Add $\frac{1}{9}$ $1/9$ to the right side and add $\frac{3}{7} (\frac{1}{9})$ $3/7(1/9)$ to the left side because we factored out $\frac{3}{7}$ $3/7$ in the beginning. This process will keep the equation balanced.

$\frac{3}{7} (\frac{1}{9}) - \frac{12}{7} = \frac{3}{7} (x^{2} - \frac{2}{3} x + \frac{1}{9})$ $color(red)(3/7(1/9))-12/7=3/7(x^2-2/3x+color(red)(1/9))$

Simply

$\frac{3}{7} (\frac{1}{93}) - \frac{12}{7} = \frac{3}{7} (x^{2} - \frac{2}{3} x + \frac{1}{9})$ $color(red)(cancel3/7(1/(cancel9 3)))-12/7=3/7(x^2-2/3x+color(red)(1/9))$

$\frac{1}{21} - \frac{12}{7} = \frac{3}{7} (x^{2} - \frac{2}{3} x + \frac{1}{9})$ $1/21-12/7=3/7(x^2-2/3x+color(red)(1/9))$

Find Common Denominator

$\frac{1}{21} - \frac{12}{7} \cdot \frac{3}{3} = \frac{3}{7} (x^{2} - \frac{2}{3} x + \frac{1}{9})$ $1/21-12/7*color(red)(3/3)=3/7(x^2-2/3x+color(red)(1/9))$

$\frac{1}{21} - \frac{36}{21} = \frac{3}{7} (x^{2} - \frac{2}{3} x + \frac{1}{9})$ $1/21-36/21=3/7(x^2-2/3x+color(red)(1/9))$

The right side is a perfect square trinomial

$\frac{1}{21} - \frac{36}{21} = \frac{3}{7} {(x - \frac{1}{3})}^{2}$ $1/21-36/21=3/7(x-1/3)^2$

$- \frac{35}{21} = \frac{3}{7} {(x - \frac{1}{3})}^{2}$ $-35/21=3/7(x-1/3)^2$

$- \frac{355}{213} = \frac{3}{7} {(x - \frac{1}{3})}^{2}$ $-(cancel35 5)/(cancel 21 3)=3/7(x-1/3)^2$

$- \frac{5}{3} = \frac{3}{7} {(x - \frac{1}{3})}^{2}$ $-5/3=3/7(x-1/3)^2$

Add $\frac{5}{3}$ $5/3$ from both sides

$\frac{5}{3} - \frac{5}{3} = \frac{3}{7} {(x - \frac{1}{3})}^{2} + \frac{5}{3}$ $color(red)(5/3)-5/3=3/7(x-1/3)^2color(red)(+5/3)$

$0 = \frac{3}{7} {(x - \frac{1}{3})}^{2} + \frac{5}{3}$ $0=3/7(x-1/3)^2color(red)(+5/3)$

Vertex form $\Rightarrow y = {(x - h)}^{2} + k$ $=> y=(x-h)^2+k$

Vertex $\Rightarrow (h, k) \Rightarrow (\frac{1}{3}, \frac{5}{3})$ $=> (h,k) => (1/3,5/3)$

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