SOLUTION: What function has the following characteristics? A zero at x = 3 A hole when x = 5 A vertical asymptote at x = -1 A horizontal asymptote at y = 3 A y-intercept at y = -2

Algebra ->  Functions -> SOLUTION: What function has the following characteristics? A zero at x = 3 A hole when x = 5 A vertical asymptote at x = -1 A horizontal asymptote at y = 3 A y-intercept at y = -2      Log On


   

Question 1158026: What function has the following characteristics?
A zero at x = 3
A hole when x = 5
A vertical asymptote at x = -1
A horizontal asymptote at y = 3
A y-intercept at y = -2
Answer by KMST(5385) About Me  (Show Source):
You can put this solution on YOUR website!
There could be other ways to get a function that does that, but we could easily find a ratio of polynomials (a rational function) that could do that.
Characteristic #1: A zero at x = 3
Having %28x-3%29 as a factor would do that.

Characteristic #2: A hole when x = 5
Having %28x-5%29%2F%28x-5%29 as a factor (having %28x-5%29 as a factor in numerator and denominator) would do that.

Characteristic #3: A vertical asymptote at x = -1
Having %28x-%28-1%29%29=%28x%2B1%29%29 as a factor only in the denominator would do that.

So far we have the building blocks for a function that has the 3 characteristics above,
f%5B1%5D%28x%29%22=%22%28x-3%29%28x-5%29%2F%28x-5%29%2F%28x%2B1%29%22=%22%28x%5E2-8x%2B15%29%2F%28x%5E2-4x-5%29 ,
with a graph that looks like this:
graph%28300%2C300%2C-14%2C6%2C-10%2C10%2C%28x%5E2-8x%2B15%29%2F%28x%5E2-4x-5%29%29 .
It even has a horizontal asymptote, but at y=1 ,
and its y-intercept is f%5B1%5D%280%29=%28-3%29%28-5%29%2F%28-5%29%2F%28%2B1%29=-3 .
The asymptote and intercept are not yet what we want.
If we add 2 , the horizontal asymptote becomes y=3, but then x=3 is not a zero.
If we include 3 as another factor, the he horizontal asymptote becomes y=3 , but then the y- intercept becomes -9 .

We need to meet the required characteristics #4 and #5 without loosing what we already have achieved.

Characteristic #4: A horizontal asymptote at y = 3
Characteristic #5: A y-intercept at y = -2
f%5B1%5D%28x%29 has a y=1 horizontal asymptote because it is the ratio of polynomials of the same degree, and when divide numerator x%5E2-8x%2B15 by denominator x%5E2-4x-5 the quotient is 1, and the remainder polynomial degree is less,
so that we could re-write f%5B1%5D%28x%29 as quotient%2Bremainder%2Fdivisor as
f%5B1%5D%28x%29%22=%22%28%28x%5E2-4x-5%29%2B%28-4x%2B20%29%29%2F%28x%5E2-4x-5%29%22=%221%2B%28-4x%2B20%29%2F%28x%5E2-4x-5%29 ,
and we know that
Of course the reason the quotient is 1 is that the leading coefficients of x%5E2-8x%2B15 are x%5E2-4x-5 both 1, and their ratio is 1 .
The leading coefficient of x%5E2-8x%2B15 is the product of the leading coefficients of the factors %28x-3%29 and %28x-5%29 that we had to include to get characteristics #1 and 2.
The leading coefficient of x%5E2-4x-5 is the product of the leading coefficients of the factors %28x-3%29 and %28x%2B1%29 that we had to include to get characteristics #2 and #3.
The value of the y-intercept %28-3%29%28-5%29%2F%28-5%29%2F%28%2B1%29=-3 depended only on the independent terms of factors %28x-3%29 , %28x-5%29 , and %28x%2B1%29 .
We could include an extra factor %28x%2B1%29 in the denominator,
and an extra factor %283x%2Bb%29 in the numerator
to try to achieve characteristics #4 and #5.
Horizontal asymptote y=3 is achieved by f%28x%29=%283x%2Bb%29%28x%5E2-8x%2B15%29%2F%28x%2B1%29%2F%28x%5E2-4x-5%29
because the ratio of leading coefficients would be 3%2F1=3 .
To get "A y-intercept at y = -2"we need f%280%29=b%2A15%2F%281%28-5%29%29=15b%2F%28-5%29%0D%0A=-3b=-2-->b=2%2F3%29%29%29%0D%0APutting+it+all+together+we+get%0D%0A%7B%7B%7Bf%28x%29%22=%22%283x%2B2%2F3%29%28x-3%29%28x-5%29%2F%28x%2B1%29%2F%28x-5%29%2F%28x%2B1%29%22=%22
%289x%2B2%29%28x-3%29%28x-5%29%2F%28x%2B1%29%5E2%2F%283x-15%29%22=%22%289x%5E3-70x%5E2-151x-30%29%2F%283x%5E3-9x%5E2-27x-15%29
The graph would then be

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