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]}
