Question

How do you find domain and range for `y= x^2-2`?

Answer 1

`x inRR,y in[-2,oo)`

Explanation:

`" y is well defined for all real values of x"`

`"domain is "x inRR`

`(-oo,oo)larrcolor(blue)"in interval notation"`

`"a quadratic in the form "y=x^2+c`

`"has a minimum turning point at "(0,c)`

`y=x^2-2" has a minimum turning point at "(0,-2)`

`"range is "y in[-2,oo)`
graph{x^2-2 [-10, 10, -5, 5]}

Answer 2

Domain: `\mathbb{R}`.

Range: `[-2,infty)`.

Explanation:

DOMAIN

This function is a polynomial, which means that it is a sum of powers of `x` with some coefficients, i.e.

`f(x)=a_0+a_1x+a_2x^2+...+a_nx^n`

This means that, for every input `x`, you must:

• compute the powers `x`, `x^2`, ..., `x^n`. This can be done with no restrictions on `x`.
• Multiply each power for its coefficient:
  `x\to a_1x`,
  `x^2\toa_2x^2`,
  ..
  `x^n\to a_nx^n`.
  Again, this can be done for every input.
• Finally, you have to sum all this pieces, and you can always sum a finite number of terms.

This proves that the domain of every polynomial is the whole set of real numbers `\mathbb{R}`, because you can perform the required calculations for every input `x \in \mathbb{R}`.

RANGE

Since this is a polynomial of degree `2`, it represents a parabola. And since the leading term `x^2` has a positive coefficient, the parabola is concave up.

This means that the parabola has a point of minimum, but it has no upper bound. Its range is thus something like `[a, \infty)`.

To find the minimum, we can either derive the parabola, or use the formula for the vertex: given a parabola `ax^2+bx+c`, the `x` coordinate of the vertex is `-b/(2a)`.

In your case, `a=1`, `b=0`, `c=-2`. The formula yields that the `x` coordinate of the vertex is `0`.

The minimum is thus the image of `0`, which is `f(0)=0^2-2 = -2`.

So, the range is `[-2,infty)`.