Posted by Esther on Wednesday, January 19, 2011 at 9:59pm.
f(x)=x^2+3 Solve for f(2)and f(5)
I need help, can someone please help me by explaining step by step how to solve this problem, I am most grateful to you. Thank you before hand.

Algebra 1  helper, Wednesday, January 19, 2011 at 10:04pm
just plug in the values for x and solve
f(x) = x^2 + 3
f(2) = 2^2 + 3
f(2) = ?
f(x) = x^2 + 3
f(5) = 5^2 + 3
f(5) = ? 
Algebra 1  Esther, Wednesday, January 19, 2011 at 10:10pm
Please forgive me, math is not my forte as you can see. Does that mean that f(5)= 5? and f(2)= 2? Thus, the answer is (5,2)or just 5 and 2? I guess what I'm trying to determing is how do you write the answer? Thank you and pleae excuse my confusion.

Algebra 1  helper, Wednesday, January 19, 2011 at 10:17pm
f(x) = x^2 + 3
f(2) = 2^2 + 3
f(2) = ?
2^2 = 4
f(2) = 2^2 + 3
f(2) = 4 + 3
f(2) = 7
the answer for f(2) is 7
when x = 2, the function value = 7
now, you do the other one 
Algebra 1  Esther, Wednesday, January 19, 2011 at 10:29pm
Thank you for your quick response, and your much needed help. is the answer to the the second problem 10? Thank you.

Algebra 1  helper, Wednesday, January 19, 2011 at 10:37pm
No, f(5) does not = 10
f(x) = x^2 + 3
f(5) = 5^2 + 3
f(5) = ?
5^2 = ?
5^2 + 3 = ?
show me how you got 10 
Algebra 1  Esther, Wednesday, January 19, 2011 at 10:40pm
I added 5 and 2 + 3. I am obviously wrong. Sorry!

Algebra 1  Esther, Wednesday, January 19, 2011 at 10:46pm
I think the answer is 5 because of the polynomial 2, I could be wrong but doesn't that amount to 5 twos, in other words, 2,2,2,2,2, which equal 5, one whole number? Sorry, i'm lost.

Algebra 1  helper, Wednesday, January 19, 2011 at 11:40pm
f(x) = x^2 + 3
f(5) = 5^2 + 3
f(5) = ?
5^2 = 5 x 5 = 25
f(5) = 5^2 + 3
f(5) = 25 + 3
f(5) = 28 
Algebra 1  MattsRiceBowl, Thursday, January 20, 2011 at 12:17am
5^2 is "5 squared."
xxxxx
xxxxx
xxxxx
xxxxx
xxxxx
(Not a perfect square there because of the text size).
How many are in 5^2? (Just count them up. It is 5 up or 5 across in every row/column).
) 
Algebra 1  Esther, Thursday, January 20, 2011 at 1:03am
Thank you. I wholeheartedly, appreciate your help. Your help does make a big difference in how to solve some of these algebra problems. Thank you, again.

Algebra 1  helper, Thursday, January 20, 2011 at 1:23am
you're very welcome :)