Question

What is the vertex form of `y=3x^2-2x-1`?

Answer 1

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

Explanation:

Given a quadratic of the form `y=ax^2+bx+c` the vertex, `(h,k)` is of the form `h=-b/(2a)` and `k` is found by substituting `h`.

`y=3x^2-2x-1` gives `h=-(-2)/(2*3)=1/3`.

To find `k` we substitute this value back in:

`k=3(1/3)^2-2(1/3)-1 = 1/3-2/3-3/3=-4/3`.

So the vertex is `(1/3,-4/3)`.

Vertex form is `y=a*(x-h)^2+k`, so for this problem:

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

Answer 2

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

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`"to obtain this form use "color(blue)"completing the square"`

`• " the coefficient of the "x^2" term must be 1"`

`rArry=3(x^2-2/3x-1/3)`

`• " add/subtract "(1/2"coefficient of x-term")^2" to"`
`x^2-2/3x`

`y=3(x^2+2(-1/3)xcolor(red)(+1/9)color(red)(-1/9)-1/3)`

`color(white)(y)=3(x-1/3)^2+3(-1/9-3/9)`

`rArry=3(x-1/3)^2-4/3larrcolor(red)"in vertex form"`

Answer 3

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

Explanation:

You must complete the square to put this quadratic into turning point form.

First, factorise out the `x^2` coefficient to get:

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

Then halve the `x` coefficient, square it, and add it and subtract it from the equation:

`y = 3(x^2 -2/3x + 1/9) - 1/3 -1`

Note that the polynomial inside the brackets is a perfect square. The extra `-1/3` has been added to maintain equality (this is equivalent to adding and subtracting `1/9`, multiplying by `3` when removing it from the brackets).

Hence:

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

From this the turning point can be found to be located at `(1/3, -4/3)`