How do you find the domain and range for #y=5/4(x+2)^2 -1 #?

1 Answer
Jul 11, 2018

Answer:

#x inRR,y in[-1,oo)#

Explanation:

#"this is a quadratic and is defined for all real values of "x#

#"domain is "x inRR#

#(-oo,oo)larrcolor(blue)"in interval notation"#

#"to find the range we require to find the vertex and if it is"#
#"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"#

#y=5/4(x+2)^2-1" is in this form"#

#"with "(h,k)=(-2,-1)larrcolor(red)"vertex"#

#"since "a>0" then minimum turning point "uuu#

#"range is "y in[-1,oo)#
graph{5/4(x+2)^2-1 [-10, 10, -5, 5]}