# How do you find the domain and range of f(x)=3x^2-5?

Apr 16, 2017

Domain = $\mathbb{R}$ = ($- \infty , \infty$)
Range = $y \ge - 5$

#### Explanation:

Domain is the possible x-values that can be put into the equation.
Range is all the possible y-values that can come out of the equation.

All quadratic equations have a domain of all real numbers, because any x-value can be plugged into the equation, and because the parabola extends width wise for infinity.

A domain of all real numbers can be written as $\mathbb{R}$ or ($- \infty , \infty$).

The range of quadratic equations depends on the vertex. To find the vertex of the equation, first put it in standard form ($a {x}^{2} + b x + c$), which it already is. Then use the formula $\frac{- b}{2 a}$.

$\frac{- b}{2 a}$ = $\frac{0}{6}$ = $0$

the x-value of the vertex is 0. substitute it back in to get the y-value.

$f \left(0\right) = 3 {\left(0\right)}^{2} - 5 = 0 - 5 = - 5$

So the vertex is (0, -5). The range is either $y \ge - 5 \text{ }$or$\text{ } y \le - 5$.

Since $a$ is positive, the parabola opens upwards. So the range has to be y is greater than or equal to something.

So the range is $y \ge - 5$.

Domain and range are easier to find when you have the graph of the equation:graph{3x^2 - 5 [-10.59, 9.41, -6.16, 3.84]}