Question

If the standard form of an equation is `y=1/2x^2-2x+5`, what is the vertex form?

Answer 1

`y=1/2(x-2)^2+3`

Explanation:

To convert from the standard form to the vertex form, you must use completing the square, and in order to use it, the value of `a` must be "`1`". In this case

`a=1/2,` `b=-2,` `c=5`

The value of `a` here is `1/2`, so we must factor out `1/2` from the expression as follows

`y=1/2x^2-2x+5 -> y=1/2(x^2-4x+10)`

The values of `a,b,c` in the expression `x^2-4x+10` are

`a=1,` `b=-4,` `c=10`

The expression `x^2-4x+10` is not a perfect square trinomial. We can make it a perfect square by finding the value of `c` which is equal to `(b/2)^2`

`color(red)(c)=(b/2)^2= ((-4)/2)^2=(-2)^2=4`

Next, we add `color(red)(c)` to the expression `x^2-4x+10`, but to keep the equation balanced, we must subtract `color(red)(c)` also, but without cancelling

`x^2-4x+10=x^2-4xcolor(red)(+4)+10color(blue)(-4)=x^2-4x+4+6`

So

`y=1/2(x^2-4x+4+6)`

Bring `6` outside from the brackets, and in order to do that you must multiply it by `1/2`. When you do that you will get

`y=1/2(x^2-4x+4)+3`

Finally, now you can factor `x^2-4x+4` into `(x-2)^2` and you get

`y=1/2(x-2)^2+3` Which is the vertex form.