# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of  y=x^2 - 3?

Dec 16, 2017

Axis of symmetry is $x = 0$ and vertex is $\left(0 , - 3\right)$. $x$-intercepts are $\left(- \sqrt{3} , 0\right)$ and $\left(\sqrt{3} , 0\right)$ and only $y$-intercept is $\left(0 , - 3\right)$. Domain is $\left(- \infty , \infty\right)$ and range is $\left[- 3 , \infty\right)$.

#### Explanation:

For an equation in the form $y = a {\left(x - h\right)}^{2} + k$, while axis of symmetry is $x - h = 0$, vertex is $\left(h , k\right)$.

As we can write $y = {x}^{2} - 3$ as $y = {\left(x - 0\right)}^{2} + \left(- 3\right)$, axis of symmetry is $x = 0$ i.e. $y$-axis and vertex is $\left(0 , - 3\right)$.

For $x$-intercept, we should put $y = 0$ i.e. ${x}^{2} = 3$ and hence $x$-intercepts are at $\left(\sqrt{3} , 0\right)$ and $\left(- \sqrt{3} , 0\right)$. For $y$-intercepts we should put $x = 0$ and only $y$-inercept is at $\left(0 , - 3\right)$.

While there are no limitations on $x$ and hence domain for $y = {x}^{2} - 3$, is $\left(- \infty , \infty\right)$. But as ${x}^{2} \ge 0$, $y$ cannot be less than $- 3$ and hence range is $\left[- 3 , \infty\right)$.

graph{ x^2-3 [-10, 10, -5, 5]}