Question

How do you find the axis of symmetry and vertex point of the function: `y = -x^2 + 6x + 2`?

Answer 1

`x=3," vertex "=(3,11)`

Explanation:

`"given the equation in "color(blue)"standard form"`

`•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0`

`"then the x-coordinate of the vertex which is also "`
`"the axis of symmetry is"`

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

`y=-x^2+6x+2" is in standard form"`

`"with "a=-1,b=6" and "c=2`

`rArrx_(color(red)"vertex")=-6/(-2)=3`

`"substitute this value into the equation for y"`

`rArry_(color(red)"vertex")=-(3)^2+6(3)+2=11`

`rArrcolor(magenta)"vertex "=(3,11)`

`"and axis of symmetry is "x=3`