How do you find the domain and range of #y=2x^2-4x+3#?

2 Answers
Jul 2, 2018

Answer:

Range: #y>=1#
Domain: #x in RR# (or "all x")

Explanation:

the domain is "all x" (#x in RR#) because it doesn't matter which x you there, you will get "some y".

in order to find the range of the function, giving that it is "a smiling parabola" (#a>0#), you just need to find the min (#x_min=-b/(2a)#):

#x_min=-b/(2a)=4/(2*2)=1#

now find #y_min=2*1^2-4*1+3=2-4+3=-2+3=1#

so the min is #(1,1)#, so the range is every y equal or above (#y>=1#)

Jul 2, 2018

Answer:

#x inRR,y in[1,oo)#

Explanation:

#"this is a polynomial of degree 2 and is defined for all real"#
#"values of "x#

#"domain is "x inRR#

#"to find the range we require the vertex and whether it is"#
#"a maximum or minimum turning point"#

#"the equation of a parabola in "color(blue)"vertex form"# is.

#•color(white)(x)y=a(x-h)^2+k#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#"to obtain this form use "color(blue)"completing the square"#

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

#color(white)(y)=2(x^2+2(-1)x color(red)(+1)color(red)(-1)+3/2)#

#color(white)(y)=2(x-1)^2+1larrcolor(blue)"in vertex form"#

#color(magenta)" vertex "=(1,1)#

#"Since "a>0" then minimum turning point " uuu#

#"range is "y in[1,oo)#
graph{2x^2-4x+3 [-10, 10, -5, 5]}