Respuesta :
f(x) = -20x²+14x+12 and g(x) = 5x-6
[tex] \frac{f(x)}{g(x)} = \frac{-20 x^{2} + 14x+12}{5x-6} [/tex]
[tex]= \frac{-(20 x^{2} -14x-12)}{5x-6} [/tex]
[tex] \frac{-[20 x^{2}-24x+10x-12] }{5x-6} [/tex]
=[tex] \frac{-[2x(10x-12)+1(10x-12)]}{5x-6} [/tex]
=[tex] \frac{-[(2x+1)(10x-12)]}{5x-6} [/tex]
=[tex] \frac{-2(2x+1)(5x-6)}{5x-6} [/tex]
=[tex]-2(2x+1)[/tex]
Now, since this is a linear equation, it is defined at every real number.
Therefore, domain is x∈(⁻∞,⁺∞)
[tex] \frac{f(x)}{g(x)} = \frac{-20 x^{2} + 14x+12}{5x-6} [/tex]
[tex]= \frac{-(20 x^{2} -14x-12)}{5x-6} [/tex]
[tex] \frac{-[20 x^{2}-24x+10x-12] }{5x-6} [/tex]
=[tex] \frac{-[2x(10x-12)+1(10x-12)]}{5x-6} [/tex]
=[tex] \frac{-[(2x+1)(10x-12)]}{5x-6} [/tex]
=[tex] \frac{-2(2x+1)(5x-6)}{5x-6} [/tex]
=[tex]-2(2x+1)[/tex]
Now, since this is a linear equation, it is defined at every real number.
Therefore, domain is x∈(⁻∞,⁺∞)
Answer:
-2(2x+1)
Domain : (-∞, 5/6) U (5/6 , ∞)
Step-by-step explanation:
f(x) = -20x^2 + 14x + 12 and g(x) = 5x- 6
[tex]f/g= \frac{f(x)}{g(x)} = \frac{-20x^{2} + 14x+12}{5x-6}[/tex]
Factor -20x^2 +14x+12, GCF is -2
-2(10x^2-7x-6)
we find two factors whose product is -60 and sum is -7
-12* 5= -60
-12+5 = -7
-2(10x^2-12x+5x-6)
-2((10x^2-12x)+(5x-6)
-2(2x(5x-6)+1(5x-6))
-2(2x+1)(5x-6)
Replace it in the numerator
[tex]\frac{-2(2x+1)(5x-6)}{5x-6}[/tex]
Cancel out 5x-6
So it becomes -2(2x+1)
To get the domain , we ignore the values of x that makes the denominator 0 in the original f(x)/ g(x)
we set denominator =0 and solve for x
5x-6=0
5x=6
divide by 5
x= 5/6
domain is all real numbers except 5/6
(-∞, 5/6) U (5/6 , ∞)
