SOLUTION: 15) f(x) = x + 4 a. f(5) = b. f(–7) = c. f(0) = 16) f(x) = 6x + 5 a. f(12) = b. f(–1/2) = c. f(0) = 17) h(x) = x2 – 4x + 5 a. h(4) = b. h(–4) = c.

Algebra ->  Expressions-with-variables -> SOLUTION: 15) f(x) = x + 4 a. f(5) = b. f(–7) = c. f(0) = 16) f(x) = 6x + 5 a. f(12) = b. f(–1/2) = c. f(0) = 17) h(x) = x2 – 4x + 5 a. h(4) = b. h(–4) = c.      Log On


   

Question 180587: 15) f(x) = x + 4
a. f(5) =
b. f(–7) =
c. f(0) =

16) f(x) = 6x + 5
a. f(12) =
b. f(–1/2) =
c. f(0) =


17) h(x) = x2 – 4x + 5
a. h(4) =
b. h(–4) =
c. h(0) =

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
how about I do number 15 and you do the rest
:
We just need to plug in the value of x which is given into every place x occurs in the function
:
15)15) f(x) = highlight%28x%29 + 4
a. f(5) =
b. f(–7) =
c. f(0) =
:
a)f(5)=highlight%285%29+4=9........see where I highlighted it that is where
:
I plugged in the given value of 5 into x+4
:
b)f(-7)=highlight%28-7%29+4=-3
:
c)f(0)=highlight%280%29+4=4
do the same thing for problem 16 and 17
:
remember in 17 there is an x squared and and x term....put the values in ALL occurences of x