Question

How do you find the vertex and the intercepts for `y = -x^2 - 4x + 7`?

Answer 1

Vertex is at `(-2,11)`, y intercept is at `(0,7)`, x-interceps are at `(-5.32 , 0) and (1.32,0)`

Explanation:

`y= -x^2-4x+7 or y = -(x^2+4x)+7 or y = -(x^2+4x +4)+4+7` or
`y = -(x+2)^2+11` comparing with standard equation `y=a(x-h)^2+k, (h,k)` being vertex we find here vertex at `h= -2 ,k=11) or (-2,11)`

y-intercept can be found by putting `x=0` in the equation, `y= 7`
y intercept is at `(0,7)`

x-intercept can be found by putting `y=0` in the equation , `-(x+2)^2 +11 =0 or (x+2)^2 =11 or (x+2) = +-sqrt 11 :. x= -2+sqrt11 ~~ 1.32 (2dp)` or `x= -2 - sqrt11 ~~ -5.32 (2dp)`
x-intercepts are at `(-5.32 , 0) and (1.32,0)` [Ans]