(x+2)(x+1) = x^2 + x + 2x + 2
simplify it
x^2 + 3x + 2
1)
The domain
is every value of x for which f(x) is a real number.
f(x) = 13 / (10-x)
The only x value that would not produce a real number for f(x) is 10, since you
cannot divide a number by zero. Answer is C
2)
F(x)
=(x-6)(x+6)/(x2 - 9)
The vertical asymptotes are x=3 and x=-3. Graph the function on a graphing
calculator to observe the behavior of the function at these points. There is
both a positive and negative vertical asymptote a both x=3 and x=-3. Keep in
mind that the denominator approaches zero at these points, and thus f(x) approaches
either positive or negative infinite, depending on whether the denominator, however small, is a positive or
negative number. Answer is B) 3, -3
3)
F(x) = (x2
+ 4x-7) / (x-7)
Although there is a vertical asymptote as x=7, there is no horizontal asymptote.
This makes sense. As X gets bigger, there is nothing to hold y back from
getting greater and greater. X2 is the dominant term, and it’s only
in the numerator. A) none
4)
(x2 +
8x -2) / (x-2)
This function is very similar in structure to the previous one. Same rules
apply. Dominant term only in the numerator means no horizontal asymptote.
A)None
5)
Our
function approaches 0 as x approaches infinite, and has a vertical asymptote at
x=2 and x=1.
Here’s an easy example: 10 / ((x-2)*(x-1)). At x=2 and x=1, there is both a
positive and negative vertical asymptote. As x approaches infinite, the
numerator is dominated by the denominator, which contains x (actually x2 ),
and thus y approaches zero.
Answer:
the answer was 38 sorry
Step-by-step explanation:
Answer:
D
Step-by-step explanation:
