Question

Standard form to vertex form??

Answer 1

Complete the square

Explanation:

We want to go from y intercept form `f(x)=ax^2+bx+c` into vertex form `f(x)=a(x-b)^2+c`

So take the example of
`f(x)=3x^2+5x+2`

We need to factorise the co-efficient out from the `x^2` and separate the `ax^2+bx` from the `c` so you can act upon them separately

`f(x)=3(x^2+5/3x) +2`

We want to follow this rule
`a^2+2ab+b^2=(a+b)^2`
or
`a^2-2ab+b^2=(a-b)^2`

We know that the `a^2=x^2` and
`2ab=5/3x` so `2b=5/3`

So we just need `b^2` and then we can collapse it down to `(a+b)^2`

so `2b=5/3` so `b=5/6` so `b^2=(5/6)^2`

Now we can add the `b^2` term into the equation remembering that the net sum of any additions to any equation/expression must be zero)

`f(x)=3(x^2+5/3 x+(5/6)^2)+2-3(5/6)^2`

Now we want to make the `a^2+2ab+b^2` into `(a+b)^2` so follow the same process as above

`f(x)=3(x+5/6)^2+72/36-3(25/36)`

Simply the equation
`f(x)=3(x+5/6)^2-3/36`

Now we have the result in standard form

Answer 2

General vertex form of a quadratic function:
`f(x) = a(x + b/(2a))^2 + f(-b/(2a))`
In this formula,
`(-b/(2a))` is the x-coordinate of the vertex
`f(-b/(2a))` is the y-coordinate of the vertex.
To proceed, first find `x = -b/(2a)`.
Next, find `f(-b/(2a))`
Example: Transform to vertex form -->
`f(x) = x^2 + 2x - 15`
x-coordinate of vertex:
`x = - b/(2a) = -2/2 = - 1`
y-coordinate of vertex:
`f(-b/(2a)) = f(-1) = 1 - 2 - 15 = - 16`
Vertex form:
`f(x) = (x + 1)^2 - 16`