# How do you find the domain and range of y=x^2+1?

Nov 15, 2017

Domain: $x \in \mathbb{R} \mathmr{and} x | \left(- \infty , \infty\right)$
Range : $y \ge 1 \mathmr{and} y | \left[1 , \infty\right)$.

#### Explanation:

$y = {x}^{2} + 1$ , Domain : Possible input value of $x$ is

any real value . Therefore Domain: $x \in \mathbb{R} \mathmr{and} x | \left(- \infty , \infty\right)$.

Range: $y = {x}^{2} + 1 \mathmr{and} y = {\left(x - 0\right)}^{2} + 1$ . Comparing with vertex

form of equation f(x) = a(x-h)^2+k ; (h,k) being vertex

we find here $h = 0 , k = 1 , a = 1 \therefore$ Vertex is at $\left(0 , 1\right)$

Since $a$ is positive the parabola opens upward and

vertex is the minimum point $x = 0 , y = 1$

So range is $y \ge 1 \mathmr{and} y | \left[1 , \infty\right)$.

Domain: $x \in \mathbb{R} \mathmr{and} x | \left(- \infty , \infty\right)$

Range : $y \ge 1 \mathmr{and} y | \left[1 , \infty\right)$.

graph{x^2+1 [-10, 10, -5, 5]}