can you help me?

Question

okay so i need a little help in algebra 1

if f(x)=2x+5 and g(x)=3x squared -1

g(a+1)

f(0)+g(3)

f(n)+g(n)

can you do at least one of these and explain it so i can try to understand this thankss a lot

Answer 1

g(a+1)
Replace x with a+1
g(x) = 3(x)^2 - 1
g(a+1) = 3(a+1)^2 - 1
g(a+1) = 3(a^2 + 2a + 1) - 1
g(a+1) = 3a^2 + 6a + 3 = 1
g(a+1) = 3a^2 + 6a + 2

f(0) + g(3)
For f(0), replace x with 0
f(x) = 2x + 5
f(0) = 2(0) + 5 = 5
For g(3), replace x with 3
g(3) = 3(3^2) - 1 = 27 - 1 = 26
Add them together
f(0) + g(3) = 5 + 26 = 31

f(n) + g(n) = 2n + 5 + 3n^2 - 1
= 3n^2 + 2n + 4

Answer 2

g(a+1)
putting a+1 in place of x
g(x) = 3(x)^2 - 1
g(a+1) = 3(a+1)^2 - 1
g(a+1) = 3(a^2 + 2a + 1) - 1
g(a+1) = 3a^2 + 6a + 3 = 1
g(a+1) = 3a^2 + 6a + 2

f(0) + g(3)
here,put 0 instead of x
f(x) = 2x + 5
f(0) = 2(0) + 5 = 5
here,put 3 instead of x
g(3) = 3(3^2) - 1 = 27 - 1 = 26
Adding,
f(0) + g(3) = 5 + 26 = 31
similarly,
f(n) + g(n) = 2n + 5 + 3n^2 - 1
= 3n^2 + 2n + 4
for all these type of problems u just need to substitute wht u want in place of variables..

Answer 3

if f(x)=2x+5 and g(x)=3x squared -1

a)
g(a+1) = 3(a+1)² - 1
g(a+1) = 3(a²+1+2a) - 1
g(a+1) = 3a²+3+6a-1
g(a+1) = 3a²+6a+2

b)
f(0)+g(3) =
(2(0)+5)+(3(3)² - 1) =
5+3(9)-1 = 31

c)
f(n)+g(n) =
(2(n)+5)+(3(n)² - 1) =
2n+5+3n²-1 = 3n²+2n+4

₢

Answer 4

most important thing to remember when doing this type of problem: just plug it in!
g(a+1) = 3(a+1)^2 -1 so you get 3(a^2 + 2a +1) -1. then distribute the 3 to get 3a^2 +6a+3 -1. so your final answer is 3a^2 +6a+2.

f(0) = 2(0) +5 = 5
g(3) = 3(3)^2 -1 = 27-1 = 26
so it's 5+26= 31

and it's basically the same concept for f(n) + g(n)