Help with this problem I don't understand it go someone please show me how to do and explain it

Question

Can someone please help me with is this problem I don't understand it

Answer the following questions for the function, g(x)=2x−12/x^2+3x−4

A) Is the point (-6,0) on the graph of y=g(x)?

1. yes
2. no

B) What is the domain of g(x)? Give your answer using interval notation.

   

C) g(−2)=    Give your answer as an integer or a completely reduced fraction.

D) For what value(s) of x does g(x)=0? x=

 

E) g(x)=−4/3. What is x? x=    Give your answers as integers or completely reduced fractions.

Answer 1

f(x) = (2x-12)/(x^2 +3x -4)

is (6,0) on the graph of f(x)?

plug in the value x=6

y = f(6) = (2(6)-12)/(36+18-4)= 0/50 = 0

UNLESS you really meant

f(x) = 2x -(12/x^2) +3x -4

then f(6) is undefined, since it involves division by zero for the 2nd term

domain is all real values, except when x^2+3x-4=0

(x+4)(x-1)=0

x=-4, x=1

domain is all real values except 1 and -4

in interval notation x is an element of (-infinity, -4)U(-4,1)U(1, infinity)

UNLESS you meant f(x)=2x -(12/x^2)+3x-4

then the domain is all real values except 0

in interval notation (-infinity,0)U(0, infinity)

g(-2) = (2(-2)-2)/((-2)^2 +3(-2)-4) = -6/(4-6-4) = -6/-6= 1

or

g(-2)= (-4-2)/(-2+4)(-2-1) =-6/2(-3) = 1

UNLESS you meant f(x)= 2x-(2/x^2) +3x-4

then g(-2)= 2(-2)-(2/4)+3(-2)-4 = -4-1/2 -6-4= -29/2

g(x)=0, when x=6

UNLESS you meant f(x)=2x-(2/x^2)+3x-4

then g(x) never =0 for any value of x

g(x)=-4/3= (2x-2)/(x^2+3x-4)

divide by 2

-2/3 = (x-1)/(x^2+3x-4)

cross multiply

-2x^2 -6x +8 = x-1

2x^2 +7x-9=0

(2x+9)(x-1) = 0

x=1,x=-9/2

or if you meant f(x)=2x-(2/x^2)+3x-4 then

-4/3=2x -(2/x^2) +3x -4

-4x^2 = 6x^3 +9x^2 -12x^2

6x^3+x^2 = 0

x^2(6x+1)=0

x=-1, 1, -1/6