How do you find the domain and range for #y=5/4(x+2)^2 -1 #?
1 Answer
Jul 11, 2018
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]}
