Question

How do you find the vertex and the intercepts for `f(x)=-x^2+2x-6`?

Answer 1

`"see explanation"`

Explanation:

`"for a parabola in standard form "f(x)=ax^2+bx+c`

`"the x-coordinate of the vertex is "`

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

`f(x)=-x^2+2x-6" is in standard form"`

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

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

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

`y_(color(red)"vertex")=-1^2+2-6=-5`

`rArrcolor(magenta)"vertex "=(1,-5)`

`color(blue)"Intercepts"`

`• " let x = 0, in the equation for y-intercept"`

`• " let y = 0, in the equation for x-intercepts"`

`x=0toy=-6larrcolor(red)" y-intercept"`

`y=0to-x^2+2x-6=0`

`"check the "color(blue)"discriminant"`

`Delta=b^2-4ac=4-24=-20`

`Delta<0rArr" equation has no real solutions"`

`"hence there are no x-intercepts"`
graph{-x^2+2x-6 [-16.02, 16.02, -8, 8.02]}