What is the vertex form of #7y=3x^2 − 2x + 12 #?

Question

What is the vertex form of #7y=3x^2 − 2x + 12   #?

Julia Adams · Accepted Answer

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

#(cancel7y)/cancel7=3/7x^2-2/7x+12/7#

Set the equation equal to zero.

#0=3/7x^2-2/7x+12/7#

Subtract #12/7# from both sides

#0color(red)(-12/7)=3/7x^2-2/7x+12/7color(red)(-12/7)#

Simplify

#color(red)(-12/7)=3/7x^2-2/7x#

Factor out #3/7#

#-12/7=3/7(x^2-2/cancel7(cancel7/3)x)#

Simplify

#-12/7=3/7(x^2-2/3x)#

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

#((-2/3)/2)^2=(-2/3*1/2)^2=(-2/6)^2=(-1/3)^2=1/9#

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

#color(red)(3/7(1/9))-12/7=3/7(x^2-2/3x+color(red)(1/9))#

Simply

#color(red)(cancel3/7(1/(cancel9 3)))-12/7=3/7(x^2-2/3x+color(red)(1/9))#

#1/21-12/7=3/7(x^2-2/3x+color(red)(1/9))#

Find Common Denominator

#1/21-12/7*color(red)(3/3)=3/7(x^2-2/3x+color(red)(1/9))#

#1/21-36/21=3/7(x^2-2/3x+color(red)(1/9))#

The right side is a perfect square trinomial

#1/21-36/21=3/7(x-1/3)^2#

#-35/21=3/7(x-1/3)^2#

#-(cancel35 5)/(cancel 21 3)=3/7(x-1/3)^2#

#-5/3=3/7(x-1/3)^2#

Add #5/3# from both sides

#color(red)(5/3)-5/3=3/7(x-1/3)^2color(red)(+5/3)#

#0=3/7(x-1/3)^2color(red)(+5/3)#

Vertex form #=> y=(x-h)^2+k#

Vertex #=> (h,k) => (1/3,5/3)#

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Nathan Axel · Answer

undefined To rewrite the equation $7y = 3x^2 - 2x + 12$ in vertex form, we need to complete the square.

First, divide both sides of the equation by 7 to isolate $y$:
$$y = \frac{3}{7}x^2 - \frac{2}{7}x + \frac{12}{7}$$

Next, complete the square for the quadratic term $x^2 - \frac{2}{7}x$. To do this, take half of the coefficient of $x$, square it, and add it inside the parentheses. The coefficient of $x$ here is $-\frac{2}{7}$, so half of it is $-\frac{1}{7}$, and when squared, it becomes $\frac{1}{49}$. Add $\frac{1}{49}$ inside the parentheses:
$$y = \frac{3}{7}\left(x^2 - \frac{2}{7}x + \frac{1}{49}\right) + \frac{12}{7} - \frac{3}{7}\left(\frac{1}{49}\right)$$

Now, factor the quadratic term inside the parentheses:
$$y = \frac{3}{7}\left(x - \frac{1}{7}\right)^2 + \frac{12}{7} - \frac{3}{343}$$

To simplify, combine the constants:
$$y = \frac{3}{7}\left(x - \frac{1}{7}\right)^2 + \frac{12 \cdot 49 - 3}{7 \cdot 49}$$
$$y = \frac{3}{7}\left(x - \frac{1}{7}\right)^2 + \frac{588 - 3}{343}$$
$$y = \frac{3}{7}\left(x - \frac{1}{7}\right)^2 + \frac{585}{343}$$

So, the vertex form of the equation $7y = 3x^2 - 2x + 12$ is $y = \frac{3}{7}\left(x - \frac{1}{7}\right)^2 + \frac{585}{343}$.