Respuesta :

PenguinHelper

To start to solve this problem, we need to know what vertex form is. The vertex form of a parabola is. The vertex form of a parabola is a(x-h) + k, where k is the vertical shift, h is the horizontal shift, and a is the value that tells the stretch.


To start to solve this equation, we want to start to create a difference of two squares.

y = 2(x²+[tex]\frac{1}{2}[/tex]x) We do this step to make the x² have a coefficient of 1

Now, we want to complete the square. To complete the square, we take 1/2 of the coefficient of x, and then square that.

1/2 * 1/2 = 1/4, and 1/4²=1/16

That means that we need to add 1/16 inside and outside the parenthesis.

We get:

y = 2(x²+1/2x + 1/16) - 1/16*2

We do -1/16*2 on the outside because since we added it inside the parenthesis, we need to take it away somewhere else (if that makes sense). The two is there because there is a two in front of the parenthesis.

We get:

y = 2(x+1/4)² - 1/8, by completing the square and simplifying, and this is the final answer.



GoddessofDork

Steps:

  • Vertex Form (aka general form): [tex]y=a(x-h)^2+k[/tex] , with (h,k) as the vertex.

So for this, we are going to be completing the square. Firstly, we have to factor out the 2 out of the right side of the equation to make the x² coefficient 1:

[tex]y=2(x^2+\frac{1}{2}x)[/tex]

Next, we want to make the quantity inside of the parentheses a perfect square. To find the constant of this soon-to-be perfect square, divide the x coefficient by 2 and then square the quotient. In this case:

[tex]\frac{1}{2}\div \frac{2}{1}\\\\\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\\\\\\(\frac{1}{4})^2=\frac{1^2}{4^2}=\frac{1}{16}[/tex]

Now, we are going to be adding 1/16 inside of the parentheses. Now we want to cancel this quantity out (note: addition property of equality). To do this, we want to add the product of 1/16 and 2 (since 2 is multiplying with 1/16 on the right side) to the left side:

[tex]\frac{1}{16}\times\frac{2}{1}=\frac{2}{16}=\frac{1}{8}\\\\y+\frac{1}{8}=2(x^2+\frac{1}{2}x+\frac{1}{16})[/tex]

Now that the quantity inside the parentheses is a perfect square, factor:

  • Tip: [tex](x+y)^2=x^2+2xy+y^2[/tex]

[tex]x^2+\frac{1}{2}x+\frac{1}{16}=(x+\frac{1}{4})^2\\\\y+\frac{1}{8}=2(x+\frac{1}{4})^2[/tex]

Lastly, subtract both sides by 1/8:

[tex]y=2(x+\frac{1}{4})^2-\frac{1}{8}[/tex]

Answer:

In short, your vertex form is: [tex]y=2(x+\frac{1}{4})^2-\frac{1}{8}[/tex]