Can someone please help me with this algebra 2 question???????????????

Question

Ok......I have to turn this in for pionts.......I did it but I am not sure I got the right answers........can someone try and figure it out so I can see if my answers are correct?????

Evaluate the functions:

f(x)= 2x-5 / 3x^2+7

g(x)= absolute value of 2x^2+1 / x+3

a) find f(-2/3) no decimals......
b) find g(-3)
c) find g(-6)
d) find f(a+2)

Thanks in advance for all of your help!!!!!!!!!!!!!!!!!!!!!

Answer 1

a ((-4/3)-5)/(3(4/9)+7) = (-19/3)/(25/3) = -19/25.
b (18+1)/(-3+3) = undefined: denominator 0.
c (72+1)/(-6+3) = -73/3.
d (2(a+2)-5)/3(a+2)^2+7) = (2a-1)/(3a^2+12a+19)..
All this assumes parenthesization of original is:
f(x) = (2x-5)/(3x^2+7). If not, appropriate changes are needed.

Answer 2

All you need to do is to substitute x by the given value.
Since you didn't put in parenthesis... I'm assuming...
f(x) = (2x-5)/3x^2 + 7
a. 9/4
b and c should be easy to find.
d. [2(a+2) - 5]/3(a + 2)^2 + 7 ---> hope this will help you determine if you have the correct step!

Answer 3

a) (2*-2/3 - 5)/(3*(-2/3)^2+7) = (-4/3-5)/(3*4/9 + 7)
= (-19/3)/(25/3) = -19/25

b) abs (2*(-3)^2 + 1)/(-3+3) is undefined (zero denominator)

c) abs((2*36+1)(9) = 73/9

d) (2(a+2)-5)/(3*(a+2)^2+7)
= (2a+4-5)/(3*(a^2+4a+4+7)

= (2a-1)(3a^2+12a+33)

Answer 4

plug the value in.

example:
B would look like this:

2 (-3)^2+1 / (-3)+3

just plug the values in () into the x position in the equations