Question

How do you find the vertex and the intercepts for `f(x) = -x^2 + 2x + 5`?

Answer 1

it vertex is maximum at `(1,6)`

x-intercepts `= 1 +- sqrt6`

y-intercept `= 5`

Explanation:

`f(x) = -x^2 + 2x + 5`

`f(x) = -(x^2 -2x) + 5`

`f(x) = -(x -1)^2 +(-1)^2 + 5`

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

since coefficient of `(x-1)` is -ve value, it vertex is maximum at `(1,6)`

to find x-intercept, plug in `f(x) =0` in the equation, therefore
`0 = -(x -1)^2 +6`
`(x -1)^2 = 6`
`x -1 = +- sqrt6`
`x = 1 +- sqrt6`

to find y-intercept, plug in `x =0` in the equation, therefore
`f(0) = -(0-1)^2 +6`
`f(0) = -( -1)^2 +6`
`f(0) = -1 +6 = 5`

Answer 2

`"see explanation"`

Explanation:

`"for the standard form of a parabola " y=ax^2+bx+c`

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

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

`"with " a=-1,b=2" and " c=5`

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

`"substitute this value into function for y-coordinate"`

`rArry_(color(red)"vertex")=-(1)^2+(2xx1)+5=6`

`rArrcolor(magenta)"vertex "=(1,6)`

`color(blue)" for intercepts"`

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

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

`x=0toy=0+0+5=5larrcolor(red)" y-intercept"`

`y=0to-x^2+2x+5=0`

`"solve using the "color(blue)"quadratic formula"`

`x=(-2+-sqrt(4+20))/(-2)=(-2+-sqrt24)/-2=(-2+-2sqrt6)/-2`

`rArrx=1+-sqrt6`

`rArrx~~3.45,x~~-1.45larrcolor(red)" x-intercepts"`
graph{-x^2+2x+5 [-12.65, 12.66, -6.34, 6.3]}