Posted by Angel Eyes on Sunday, August 22, 2010 at 1:38pm.
f(x)= 4(x+5)^2+3
The vertex is : 5,3
The line of symmetry is x= 3
The maximum/minimum value of f(x)= 5
Is the value of f(5)=3, a minimum or maximum? Minimum
Graphing would open from the bottom going up on the negative side.

Math/Algebra...can you check this please...  Reiny, Sunday, August 22, 2010 at 1:49pm
write your vertex as (5,3)
the line of symmetry is the x of the vertex, so
line of symmetry is x = 5
the max of the function is 3, (the y value of the vertex)
since the value of "a" of your function is negative, the parabola opens downwards, thus a Maximum.

Math/Algebra...can you check this please...  Angel Eyes, Sunday, August 22, 2010 at 1:54pm
Okay. I had the vertex written in correctly. the line of symmetry & max function I had backwards for some unknown reason. The other 2 parts I had totally wrong.
Thank you for checking this for me Reiny! I was leery at posting it until I had it checked. Good thing I did!

Math/Algebra...can you check this please...  Angel Eyes, Sunday, August 22, 2010 at 2:04pm
I put the vertex as (5,3) and it is saying it was wrong...ughhhhhhhh.

Math/Algebra...can you check this please...  bobpursley, Sunday, August 22, 2010 at 2:27pm
the vertex is indeed (5,3)

Math/Algebra...can you check this please...  MathMate, Sunday, August 22, 2010 at 5:54pm
For
f(x)=a(xh)+k, the vertex is at:
(h,k).
Since the function is already in that form, where
a=4,
h=5, and
k=3
f(x)= 4(x+5)^2+3
as Mr. Pursley stated:
"the vertex is indeed at (5,3)".
Reread the question to see if
1. there are errors of transcription, especially signs
2. there are instructions on the format of presentation of answers.
There are times (but rare) that the computer rejects the correct answers. Hope this is not one of them.
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