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

2 Answers
Jul 2, 2018

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

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]}