MalakWael1363 MalakWael1363
  • 19-06-2020
  • Mathematics
contestada

Rewrite the function by completing the square. h(x)=2x^2+11x+15


h(x)=__(x+__)^2+__

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Alfpfeu
Alfpfeu Alfpfeu
  • 19-06-2020

Hello,

[tex]h(x) = 2x^2+11x+15[/tex]

we can notice that

[tex](x+\dfrac{11}{4})^2 = x^2+\dfrac{11}{2}x+{(\dfrac{11}{4})}^2[/tex]

so we can write

[tex]2x^2+11x = 2 [ {(x+\dfrac{11}{4})^2-{(\dfrac{11}{4})}^2 ][/tex]

then it comes

[tex]h(x) = 2 [ {(x+\dfrac{11}{4})^2-{(\dfrac{11}{4})}^2 ] + 15\\= 2 {(x+\dfrac{11}{4})^2 +15-{\dfrac{11^2}{8}}\\= 2 {(x+\dfrac{11}{4})^2 +{\dfrac{15*8-11^2}{8}}[/tex]so

[tex]h(x) = 2{(x+\dfrac{11}{4})^2-{\dfrac{1}{8}}[/tex]

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