Question

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

Answer 1

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`)

Answer 2

`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]}