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

2 Answers
Jun 26, 2018

Answer:

#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]}

Jun 26, 2018

Answer:

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)#.