Question

How do you write the quadratic function `y=-2x^2+6x-3` in vertex form?

Answer 1

The vertex form of a quadratic of the form `y = ax^2+bx+c` is:

`y = a(x-h)^2+k" [1]"`

where `h = -b/(2a)` and `k = ah^2+bh+c`

Explanation:

Given: `y=-2x^2+6x-3`

Please observe that `a =-2, b = 6 and c = -3`

Substitute the value of `a = -2` into equation [1]:

`y = -2(x-h)^2+k" [2]"`

Compute the value of h:

`h = -6/(2(-2))`

`h = 3/2`

Substitute the value of `h = 3/2` into equation [2]:

`y = -2(x-3/2)^2+k" [3]"`

Compute the value of k:

`k = -2(3/2)^2 + 6(3/2)-3`

`k = 3/2`

Substitute the value of `k = 3/2` into equation [3]:

`y = -2(x-3/2)^2+3/2 larr` this is the vertex form

Answer 2

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

Explanation:

Vertex form of quadratic equation `ax^2 + bx + c` is
`y = a(x - x_s)^2 + y_s`

`x_s = -b / (2a)`

`y_s = -(b^2/4a) + c`

Given equation is
`y = -2x^2 + 6x -3`
`a = -2, b = 6, c = -3`

`x_s =( -6) / (2*(-2)) = 3/2`
`y_s = -b^2/(4a) + c =-( -6^2) / (4 * -2) + (-3) = -(-9/2) - 3 = 3/2`

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