Question

How do you find the vertex and the intercepts for `f(x) =(x+5)^2 -1`?

Answer 1

`(-5,-1)" and "x=-6,x=-4`

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"`

`f(x)=(x+5)^2-1" is in vertex form"`

`"with "h=-5" and "k=-1`

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

`color(blue)"Intercepts"`

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

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

`x=0toy=(5)^2-1=24larrcolor(red)"y-intercept"`

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

`rArr(x+5)^2=1`

`color(blue)"take the square root of both sides"`

`sqrt((x+5)^2)=+-sqrt1larrcolor(blue)"note plus or minus"`

`rArrx+5=+-1`

`"subtract 5 from both sides"`

`rArrx=-5+-1`

`x=-5-1=-6,x=-5+1=-4larrcolor(red)"x-intercepts"`