anonymous
  • anonymous
For y = x2 + 6x - 16, Determine if the parabola opens up or down. State if the vertex will be a maximum or minimum. Find the vertex. Find the x-intercepts. Describe the graph of the equation.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
up
anonymous
  • anonymous
as coeffi. of x^2 is positive , it opens up. hence the vertex will be minimum. vertex(-b/2a , -D/4a) x-int. ,put y=0.
anonymous
  • anonymous
The Parabola opens up because the term with X^2 is positive If it was negative then it would open down.

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anonymous
  • anonymous
Yep, you're absolutely correct! Try plotting a few equations on here: http://www.mathsisfun.com/data/function-grapher.php You should get the hang of it in no time.
KenLJW
  • KenLJW
y = x2 + 6x - 16 y = (x + 3)^2 - 9 - 16 y = (x + 3)^2 - 25 notice for all values of x > 2 y is positive therefore it must open up

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