Pixlemania14 Pixlemania14
  • 25-05-2017
  • Mathematics
contestada

What is the maximum or minimum of h(t)=2x^2+4x+7 SHOW WORK FOR POINTS PLEASE

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apologiabiology
apologiabiology apologiabiology
  • 25-05-2017
that is the vertex
since the leading coefient (the term in front of the the x^2 term)

get into form
y=a(x-h)^2+k
te vertex is (h,k)
max value is k which occurs at x=h


complete the square
h(t)=(2x^2+4x)+7
h(t)=2(x^2+2x)+7
h(t)=2(x^2+2x+1-1)+7
h(t)=2((x+1)^2-1)+7
h(t)=2(x+1)^2-2+7
h(t)=2(x+1)^2+5
vertex is at (-1,5)
the max value is 5
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