How do you find the domain and range of the function `y = −2x^2 − 6x + 1` without graphing?

For the domain, I ask myself what real numbers I could use for `x` and get a real number for `y`. There's no division by `x` , so no danger of dividing by `0`. And there are no even roots, so I there's no chance of getting an imaginary answer.

The domain is all real numbers. `(-oo. oo)`

To find the range (without thinking about the graph) I need to know what real numbers could I use for `y` and get a real number for `x`

To do that, I'll sole the equation for `x` is terms of `y`.

Solve for `x`:
`y=-2x^2-6x+1`

`2x^2+6x+y-1=0`

Use the quadratic formula with `a=2`, `b=6`, `c=y-1`

`x=(-6 +-sqrt ((6)^2-4 (2)(y-1)))/4`

This will give real numbers for `x` if `36-8y+8 >= 0`

`44>=8y`

`y <= 11/2`

The range is all reals less than or equal to `11/2`.
The interval `(-oo, 5/2)`

`y=-2x^2-6x+1`

Complete the square:

`y=-2(x^2+3x )+1`

(If I knew how, I'd leave some space between "`3x`" and the "`")"`" )

`y=-2(x^2+3x +(9/4)-(9/4) )+1`

`y=-2(x^2+3x +9/4)+ 9/2 +1`

`y=-2(x+3/2)^2+ 11/2`

Now for every real value for `x`, we get a real number for `y`, so the domain is `(-oo, oo)` (All real numbers.)

Furthermore, for `x=-3/2`, the first term on the right, `-2(x+3/2)^2` is `0`, so we get `y=11/2`.

Finally, for all real `x`, the term `-2(x+3/2)^2<=0` so we will be getting `y` values less than or equal to `11/2`.

The range is `(-oo, 11/2]`