Question

What is the vertex of `y=-2x^2 + 2x + 5`?

Answer 1

`(1/2,11/2)`

Explanation:

`"given the equation of a parabola in standard form"`

`"that is "y=ax^2+bx+c`

`"then "x_(color(red)"vertex")=-b/(2a)`

`y=-2x^2+2x+5" is in standard form"`

`"with "a=-2,b=+2,c=5`

`rArrx_(color(red)"vertex")=-2/(-4)=1/2`

`"substitute this value into the equation for the corresponding"`
`"y-coordinate"`

<`rArry_(color(red)"vertex")=-2(1/2)^2+2(1/2)+5=11/2`

`rArrcolor(magenta)"vertex "=(1/2,11/2)`

Answer 2

Vertex is at `(1/2, 11/2)`.

Explanation:

The axis of symmetry is also the x value of the vertex. So we can use the formula `x=(-b)/(2a)` to find the axis of symmetry.

`x=(-(2))/(2(-2))`

`x=1/2`

Substitute `x=1/2` back into the original equation for the y value.

`y = -2(1/2)^2 + 2(1/2) + 5`

`y = 11/2`

Therefore, the vertex is at `(1/2, 11/2)`.