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

1 Answer
Apr 7, 2017

Answer:

#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]}