Question

How do you graph the parabola `y= (x+4)^2+4` using vertex, intercepts and additional points?

Answer 1

Vertex: `(-4,4)`, axis of symmetry: `x=-4`, y intercept:
`(0,20)`, additional points : `(-2,8),(-6,8),(-8,20)` opens upward.

Explanation:

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

This is vertex form of equation ,`y=a(x-h)^2+k ; (h,k)`

being vertex , here `h=-4 ,k=4,a=1`

Since `a` is positive, parabola opens upward.

Therefore vertex is at `(-4, 4)`

Axis of symmetry is `x= h or x =-4 ;`

x-intercept is found by putting `y=0` in the equation

`0=(x+4)^2+4 or or (x+4)^2=-4` or

`(x+4)^2=(2i)^2 or (x+4)= +- 2i or x= -4 +- 2i`

Since the roots are complex, parabola will have no x intercept.

y-intercept is found by putting `x=0` in the equation

`y=(x+4)^2+4 or y=20 or (0,20)`

Additional points:

`x=-2 :.y=8 , x=-6 :. y=8 , x=-8 :. y=20` or

`(-2,8) , (-6,8) and (-8,20)`

graph{(x+4)^2+4 [-38.95, 38.95, -19.47, 19.47]} [Ans]