Question

What is the axis of symmetry and vertex for the graph `y= -x^2 + 6x - 2`?

Answer 1

Vetex is at `(3, 7)` and axis of symmetry is `x = 3 ;`

Explanation:

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

`y=-(x^2-6x+3^2)+9 -2` or

`y=-(x-3)^2 + 7` . This is vertex form of equation

`y=a(x-h)^2+k ; (h,k)` being vertex , here `h=3 ,k=7`

Therefore vetex is at `(h,k) or (3, 7)`

Axis of symmetry is `x= h or x = 3 ;`

graph{-x^2+6x-2 [-20, 20, -10, 10]} [Ans]

Answer 2

`x=3" and "(3,7)`

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"`

`• " if "a>0" then graph opens up"`

`• " if "a<0" then graph opens down"`

`"express y in vertex form using the method of "color(blue)"completing the square"`

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

`rArry=-1(x^2-6x+2)`

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

`rArry=-(x^2-6xcolor(red)(+9)color(red)(-9)+2)`

`color(white)(rArry)=-(x-3)^2+7larrcolor(red)"in vertex form"`

`rArrcolor(magenta)"vertex "=(3,7)`

`"since "a<0" then parabola is vertical and opens down"`

`"the axis of symmetry is vertical and passes through the"`
`"vertex with equation "x=3`
graph{(y+x^2-6x+2)(y-1000x+3000)((x-3)^2+(y-7)^2-0.05)=0 [-20, 20, -10, 10]}