Question

How do you find the vertex of the parabola `y=-2x^2 + 12x - 13`?

Answer 1

`"vertex "=(3,5)`

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 the method of "color(blue)"completing the square"`

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

`rArry=-2(x^2-6x+13/2)`

`• " add/subtract "(1/2"coefficient of the x-term")^2" to"`
`x^2-6x`

`y=-2(x^2+2(-3)xcolor(red)(+9)color(red)(-9)+13/2)`

`color(white)(y)=-2(x-3)^2-2(-9+13/2)`

`color(white)(y)=-2(x-3)^2+5larrcolor(red)"in vertex form"`

`rArrcolor(magenta)"vertex "=(3,5)`
graph{(y+2x^2-12x+13)((x-3)^2+(y-5)^2-0.04)=0 [-10, 10, -5, 5]}

Answer 2

Vertex is at `(3,5)` .

Explanation:

`y = -2x^2+12x-13 or y = -2(x^2-6x)-13` or

`y = -2(x^2-6x +9)+18-13` or

`f(x) = -2(x-3)^2+5`. Comparing with vertex form of

equation `y = a(x-h)^2+k ; (h,k)` being vertex we find

here `h=3 , k=5 :.` Vertex is at `(3,5)` .

graph{-2x^2+12x-13 [-17.78, 17.78, -8.89, 8.89]} [Ans]