hba
  • hba
Help me with this question. Screenshot attached.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hba
  • hba
hba
  • hba
open your eyes.
hba
  • hba
I know you are.

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anonymous
  • anonymous
To clarify, the leftmost figure doesn't apply here, right? Just the one on the right?
anonymous
  • anonymous
Assuming that's the case... You're given a vertex of \(\left(\dfrac{b}{2},h\right)\), which means the equation of the parabola, in vertex form, is \[y=a\left(x-\frac{b}{2}\right)^2+h\] To solve for \(a\), you have to use the given points \((b,0)\). You have \[0=a\left(b-\frac{b}{2}\right)^2+h\\ 0=\frac{ab^2}{4}+h\\ a=-\frac{4h}{b^2}\] So your parabola's equation is \[y=-\frac{4h}{b^2}\left(x-\frac{b}{2}\right)^2+h\] I'm not sure what class this is for, so I don't know what method you have to use for finding the area.

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