How do you find the domain and range of the function #y = −2x^2 − 6x + 1# without graphing?

2 Answers
Mar 26, 2015

For the domain, I ask myself what real numbers I could use for #x# and get a real number for #y#. There's no division by #x# , so no danger of dividing by #0#. And there are no even roots, so I there's no chance of getting an imaginary answer.

The domain is all real numbers. #(-oo. oo)#

To find the range (without thinking about the graph) I need to know what real numbers could I use for #y# and get a real number for #x#

To do that, I'll sole the equation for #x# is terms of #y#.

Solve for #x#:
#y=-2x^2-6x+1#

#2x^2+6x+y-1=0#

Use the quadratic formula with #a=2#, #b=6#, #c=y-1#

#x=(-6 +-sqrt ((6)^2-4 (2)(y-1)))/4#

This will give real numbers for #x# if #36-8y+8 >= 0#

#44>=8y#

#y <= 11/2#

The range is all reals less than or equal to #11/2#.
The interval #(-oo, 5/2)#

Mar 26, 2015

#y=-2x^2-6x+1#

Complete the square:

#y=-2(x^2+3x )+1#

(If I knew how, I'd leave some space between "#3x#" and the "#")"#" )

#y=-2(x^2+3x +(9/4)-(9/4) )+1#

#y=-2(x^2+3x +9/4)+ 9/2 +1#

#y=-2(x+3/2)^2+ 11/2#

Now for every real value for #x#, we get a real number for #y#, so the domain is #(-oo, oo)# (All real numbers.)

Furthermore, for #x=-3/2#, the first term on the right, #-2(x+3/2)^2# is #0#, so we get #y=11/2#.

Finally, for all real #x#, the term #-2(x+3/2)^2<=0# so we will be getting #y# values less than or equal to #11/2#.

The range is #(-oo, 11/2]#