Answer:
x value of vertical asymptote and y value of horizontal asymptote
Step-by-step explanation:
The graph of 1/x approaches infinity as x approaches 0 (the vertical asymptote)
As x gets either bigger or smaller, 1/x approaches the x-axis (from above on the positive side, from below on the negative side) (the horizontal asymptote)
Consider 1/(x-5) + 2, at what value of x does the graph 'go nuts' ?
When the bottom of the fraction becomes 0, x - 5 becomes 0 when x = 5, so the vertical asymptote of g(x) is at x=5
What value of y does f(x) approach as x gets more positive or more negative - as x gets bigger (as an example), y approaches 0
What y value does g(x) approach as x gets bigger? Well, as x gets big, 1/(x-5) gets small, approaching 0. The smallest 0 + 2 can get is 2, so y=2 is the horizontal asymptote
It give you two point, (2,4)(-1, -1).
f(x)=mx+b
2m+b=4
-m+b=-1
2m+b+m-b=5
3m=5
m=5/3
b=2/3
f(x)=5/3x+2/3
(x+2)(x-16)
A = X
B= 2
C= X
D= -16
Answer:
0 (zero)
Step-by-step explanation:
Plug in the variable for -4 and solve
y= 2(-4)2-4(-4)
y= -8(2)-4(-4)
y=-16+16
y=0
