Question

How do you find the domain and range of `f(x)=8x^2-5x+2`?

Answer 1

`x inRR`
`y inRR,y>=39/32`

Explanation:

f(x) has no `color(blue)"excluded values".`That is values that make the function `color(blue)"undefined"` and so all real values of x can be accepted by f(x)

`rArr"domain is "x inRR`

To find the values of y in the range we require to find the vertex.

`8x^2-5x+2`

`"has " a=8,b=-5" and " c=2`

`x_(color(red)"vertex")=-b/(2a)=-(-5)/16=5/16`

To find the corresponding y-coordinate substitute this value into the equation.

`y_(color(red)"vertex")=8(5/16)^2-5(5/16)+2`

`color(white)(xxxx)=25/32-50/32+64/32=39/32`

`rArr"vertex "=(5/16,39/32)`

`rArr"range is "y inRR,y>=39/32`
graph{8x^2-5x+2 [-12.66, 12.65, -6.33, 6.33]}