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Remember that you cannot divide by 0.
An asymptote is a straight line which a functions approaches but never intersects. Is that the case here?
Factor the expression in the numerator.
Graph:
Type of Function: Linear, see solution.
Relationship: Linear
f(0.9)=5.9
f(1.1)=6.1
Graph:
Is there an asymptote at x=1? No, see solution.
f(x)=x+5
Let's first calculate the function's values for the given values of x.
| x | x^2+4x-5/x-1 | f(x) |
|---|---|---|
| - 2 | ( - 2)^2+4( - 2)-5/- 2-1 | 3 |
| - 1 | ( - 1)^2+4( - 1)-5/- 1-1 | 4 |
| 0 | ( 0)^2+4( 0)-5/0-1 | 5 |
| 1 | ( 1)^2+4( 1)-5/1-1 | Undefined |
| 2 | ( 2)^2+4( 2)-5/2-1 | 7 |
| 3 | ( 3)^2+4( 3)-5/3-1 | 8 |
From Part A, we see that the function is a straight line which means there is a linear relationship between x and y. Let's also calculate f(0.9) and f(1.1)
| x | x^2+4x-5/x-1 | f(x) |
|---|---|---|
| 0.9 | ( 0.9)^2+4( 0.9)-5/0.9-1 | 5.9 |
| 1.1 | ( 1.1)^2+4( 1.1)-5/1.1-1 | 6.1 |
Finally, we will add these points to the graph from Part A.
Notice that there is no asymptote at x=1. An asymptote is a straight line which a function approaches but never intersects. This is not the case with our function, as the function approaches a point and not a line.
To simplify the formula we have to rewrite the numerator by factoring the expression. If the expression can be factored, we should find two terms, a and b, whose sum equals the coefficient to x and whose product equals the constant.
cccll
x^2 & + & 4x &+& -5 [0.3em]
x^2 & + & (a+b)x & +& ab
x^2+4x-5= (x-1)(x+5)
a/b=.a /(x-1)./.b /(x-1).
The formula simplifies to f(x)=x+5. which means our conjecture in Part B was correct. There is in fact a linear relationship between x and y.