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Posted by **Sammy** on Friday, January 7, 2011 at 7:02pm.

f(x) = 2(x-5)^2 + 7

F(x) = -3(x+5)^2 + 7

f(x) = 2(x+5)^2 + 7

f(x) = 2(x-5)^2 -7

I think it is 2(x+5)^2 + 7, is that right

- Algebra II -
**Reiny**, Friday, January 7, 2011 at 7:08pmYes, if you are aiming for the general quadratic in the form

f(x) = a(x-h)^2 + k

- Algebra II -
**drwls**, Friday, January 7, 2011 at 7:09pmSince you do not defined what a, h and k are in terms of any function, I have no idea.

- Algebra II -
**drwls**, Friday, January 7, 2011 at 7:10pmThanks Reiny. I had never seen that general form before

- Algebra II -
**Reiny**, Friday, January 7, 2011 at 7:15pmSome authors call it the "standard" form of a parabola, it displays the vertex (h,k).

A second point on the graph is needed to find a.

- Algebra II-Please clarify one thing for me -
**Sammy**, Friday, January 7, 2011 at 7:15pmThanks Reiny-I really appreciate it-I get confused about taking the "h" at times

So if they say h= (-5), then in the equation when it is written,it is positive and vice-versa, correct?

- Algebra II -
**Reiny**, Friday, January 7, 2011 at 7:17pmcorrect, since in the formula you have

...(x-h)^2 ....

for x-(-5) it would become x+5

for x-(+5) it would become x-5

- Algebra II-Thank you -
**Sammy**, Friday, January 7, 2011 at 7:21pmThanks alot-At least I get one concept!

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