Question

How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of `y = [x^2 + 21x + 22]/16`?

Answer 1

Vertex is `(-21/2,-353/64)` and axis of symmetry is `x+353/64=0`
`y`-intercept is `11/8` and `x`-intercepts are `-1.1` and `-19.9`.
Domain is `(-oo,oo)`, range is `[-353,64,oo)`

Explanation:

The vertex form of quadratic equation is `y=a(x-h)^2+k` and its vertex is `(h,k)` and axis of symmetry is `x-h=0`.

Hence we should first convert `y=(x^2+21x+22)/16` into vertex form.

`y=(x^2+21x+22)/16`

= `1/16(x^2+2xx21/2xx x+(21/2)^2-(21/2)^2+22)`

= `1/16(x+21/2)^2-441/64+22/16`

= `1/16(x+21/2)^2-(441-88)/64`

= `1/16(x-(-21/2))^2-353/64`

Therefore, its vertex is `(-21/2,-353/64)` and axis of symmetry is `x-(-353/64)=0` or `x+353/64=0`.

`y`-intercept can be found by putting `x=0` and `x`-intercept can be found by putting `y=0`.

Hence `y`-intercept is `22/16=11/8`

and `x`-intercept are given by using quadratic formula on `x^2+21x+22=0` i.e. `x=(-21+-sqrt(21^2-4xx22))/2=(-21+-sqrt353)/2` i.e. `-1.1` and `-19.9`.

While domain is `(-oo,oo)`, range is `[-353,64,oo)`
graph{(y-(x^2+21x+22)/16)=0 [-27.33, 12.67, -10.16, 9.84]}