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Given: color(red)(y=f(x)=(x+5)^2+8y=f(x)=(x+5)2+8
The Vertex Form of a quadratic function is:
color(green)(y=f(x)=a(x-h)^2+ky=f(x)=a(x−h)2+k, where
(h,k)(h,k) is the Vertex.
Note: color(blue)(a=1, h=-5, k=8a=1,h=−5,k=8
color(green)("Step 1: "Step 1:
Vertex is at : color(blue)((-5,8)(−5,8)
Note :
If a>0a>0, then the Vertex is a Minimum Value.
Since, a=1a=1, the Vertex is color(blue)("Minimum at "(-5,8)Minimum at (−5,8)
"Range :"f(x)>= 8Range :f(x)≥8
Using Interval Notation : [8, oo)[8,∞)
color(green)("Step 2: "Step 2:
Find Domain :
Domain of f(x)f(x) is the set of all input values for which the given function is real and well-defined.
color(blue)(f(x) = (x+5)^2+8f(x)=(x+5)2+8 does not have any undefined points.
The function does not have any domain constraints.
Therefore domain is given by
color(blue)(-oo < x < oo−∞<x<∞
Using Interval Notation :
color(blue)((-oo , oo)(−∞,∞)
Hence, the required solutions are:
Domain: color(blue)([-oo < x < oo ][−∞<x<∞]
Using Interval Notation: color(blue)((-oo,oo)(−∞,∞)
Range: color(blue)([f(x)>=8 ][f(x)≥8]
Using Interval Notation: color(blue)([8,oo)[8,∞)
color(green)("Step 3: "Step 3:
Draw the graph of color(red)(y=f(x)=(x+5)^2+8y=f(x)=(x+5)2+8 to verify the solutions:
Hope it helps.