Answer:
Vertical asymptote at x = 8
Step-by-step explanation:
The given function can be rewritten as
x^2 - 64
f(x) = --------------
8(x - 8)
Even tho' this reduces to f(x) = x + 8, the function has a vertical asymptote at x = +8 because it becomes undefined there.
Answer:
(0,-2) (2/7,0)
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given is a function as 
Equating to 0 we have equation
If the function f(x) has x intercepts then the solutions are real
Let us use remainder theorem and change of signs rule
f(0) = 1>0
f(-1) = -1+3+1=3
f(-2) = -32+6+1<0
This implies there is a real root between -1 and -2.
f(1) = -1
Since f(0) and f(1) have different signs, there exists a real root between 0 and 1.
f(2) = 32-6+1>0
Since f(1) and f(2) have different signs there exists a real root between 1 and 2.
Thus there are definitely three real solutions as
one between -1 and 0, one between 0 and 1, and third between 1 and 2.
Hello,
the transformation that applies (x,y) to (-x,-y) is a symmetry by the origin.
Every symmetry transforms a line into a line.
So the new points are also collinear.
Answer:
55
Step-by-step explanation:
Let x represent the middle integer. Then the smallest is x-2 and the largest is x+2. Your requirement is that ...
(x-2)/(x+2) > 2/3
3x -6 > 2x +4 . . . . cross multiply
x > 10 . . . . . . . . . . .add 6-2x
The smallest integer satisfying this requirement is x=11. The sum of the 5 integers is 5x = 55.
The smallest sum is 55.