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

Question

Ethan Adkins · Accepted Answer

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

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#)

Genesis Allen · Answer

#x inRR,y in[1,oo)# #"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]}

Jameson Allen · Answer

undefined To find the domain, consider the quadratic expression under the square root in the quadratic formula, ensuring it's non-negative. For the range, determine the vertex of the parabola using the formula x = -b/(2a), and substitute it back into the original equation to find the corresponding y-coordinate. The range is the set of all possible y-values.