Turnerrocks12 Turnerrocks12
  • 12-09-2022
  • Mathematics
contestada

Range of hx = 2r^2 + 20r +42

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jaisurajk jaisurajk
  • 12-09-2022

Answer: Range: hx>=-8

Step-by-step explanation:

Find the vertex first:

2r^2 + 20r +42=

2(r^2 + 20r/2 +42/2)=

2(r^2 + 10r +21)=

2(r^2 + 10r +25-4)=

2(r^2 + 5r+5r +25-4)=

2(r(r+5)+r(r+5)-4)=

2((r+5)^2-4)=

2(r+5)^2-4*2=2(r+5)^2-8   ==> Vertex: (-(5), -8) ==> (-5, -8)

-8 is when the range starts

Range: hx>=-8

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