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

Jun 26, 2018

$x \in \mathbb{R} , y \in \left[- 2 , \infty\right)$

Explanation:

$\text{ y is well defined for all real values of x}$

$\text{domain is } x \in \mathbb{R}$

$\left(- \infty , \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$

$\text{a quadratic in the form } y = {x}^{2} + c$

$\text{has a minimum turning point at } \left(0 , c\right)$

$y = {x}^{2} - 2 \text{ has a minimum turning point at } \left(0 , - 2\right)$

$\text{range is } y \in \left[- 2 , \infty\right)$
graph{x^2-2 [-10, 10, -5, 5]}

Jun 26, 2018

Domain: $\setminus m a t h \boldsymbol{R}$.

Range: $\left[- 2 , \infty\right)$.

Explanation:

DOMAIN

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

$f \left(x\right) = {a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + \ldots + {a}_{n} {x}^{n}$

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:
$x \setminus \to {a}_{1} x$,
${x}^{2} \setminus \to {a}_{2} {x}^{2}$,
..
${x}^{n} \setminus \to {a}_{n} {x}^{n}$.
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 $\setminus m a t h \boldsymbol{R}$, because you can perform the required calculations for every input $x \setminus \in \setminus m a t h \boldsymbol{R}$.

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 $\left[a , \setminus \infty\right)$.

To find the minimum, we can either derive the parabola, or use the formula for the vertex: given a parabola $a {x}^{2} + b x + c$, the $x$ coordinate of the vertex is $- \frac{b}{2 a}$.

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 $f \left(0\right) = {0}^{2} - 2 = - 2$.

So, the range is $\left[- 2 , \infty\right)$.