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

1 Answer
Apr 16, 2017

Answer:

Domain = #RR# = (#-oo, oo#)
Range = #y>=-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 #RR# or (#-oo, oo#).

The range of quadratic equations depends on the vertex. To find the vertex of the equation, first put it in standard form (#ax^2 + bx + c#), which it already is. Then use the formula #(-b)/(2a)#.

#(-b)/(2a)# = #(0)/(6)# = #0#

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

#f(0) = 3(0)^2 - 5 = 0 - 5 = -5#

So the vertex is (0, -5). The range is either #y>=-5" "#or#" "y<=-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>=-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]}