Question

How do you find the vertex and the intercepts for `f(x)=-x^2+2x+4`?

Answer 1

`"vertex "=(1,5),x=1+-sqrt5`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`•color(white)(x)y=a(x-h)^2+k`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`"to obtain this form "color(blue)"complete the square"`

`y=1(x^2+2(-1)x+1-1-4)`

`color(white)(y)=-(x-1)^2+5`

`color(magenta)"vertex "=(1,5)`

`"let x = 0 for y-intercept"`

`y=-1+5=4larrcolor(red)"y-intercept"`

`"let y = 0 for x-intercepts"`

`-(x-1)^2+5=0`

`(x-1)^2=5`

`x-1=+-sqrt5`

`x=1+-sqrt5larrcolor(red)"exact values"`