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
0 or 1 or more than one for zero functions
Answer:
I am not exactly sure but it is around 170p
Step-by-step explanation:
For the answer to the question above asking the explicit equation and domain for an arithmetic sequence with the first term of 5 and the second term of 2? I think the answer for this is <span>5-2(n-1); all integers where n≥0. I hope this helps.</span>
Answer:
y=1/3x+5/3
Step-by-step explanation:
use the equation y=mx+b plug in slope for m, -2 for x and 1 for y
