SOLUTION: Write an equation for a rational function with: Vertical asymptotes at x = -6 and x = -4 x intercepts at x = -1 and x = -3 Horizontal asymptote at y = 7

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Question 1160127: Write an equation for a rational function with:
Vertical asymptotes at x = -6 and x = -4
x intercepts at x = -1 and x = -3
Horizontal asymptote at y = 7

Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


All the parts of this problem are straightforward, and the reasoning makes sense.

Memorize the rules about what makes x-intercepts and vertical asymptotes; but also understand why the rules are what they are.

An x-intercept means the graph crosses (or touches) the x-axis. That means the y value, which is the function value, is 0. In a rational function, the function value is zero whenever the numerator is 0 (unless the denominator is also zero for the same x value).

This function has x-intercepts at x=-1 and x=-3. That means the numerator of the rational function must contain factors of x+1 and x+3. (The factor x+1 means the numerator will be 0 when x = -1; the factor x+3 means the numerator will be 0 when x = -3.)

A vertical asymptote means the function is undefined for that value of x. In a rational function, that means there is a factor in the denominator that is zero for that value of x.

This function has vertical asymptotes at x = -6 and x = -4; that means the denominator has factors of (x+6) and (x+4).

At this point we have satisfied the requirements for the x-intercepts and vertical asymptotes; our function is of the form

%28%28x%2B3%29%28x%2B1%29%29%2F%28%28x%2B4%29%28x%2B6%29%29

A horizontal asymptote means that for very large positive or very large negative values the function approaches a constant value. In the rational function we have at this point, for very large x values the "x" terms dominate (the constants become insignificant), and the function value approaches x%5E2%2Fx%5E2+=+1, so the horizontal asymptote is y=1. To get a horizontal asymptote of y=7, we simply add a constant factor of 7 to the numerator.

The function we are looking for is then

%287%28x%2B3%29%28x%2B1%29%29%2F%28%28x%2B4%29%28x%2B6%29%29

Here is a graph (window -10,2,-5,5) showing the two x-intercepts:



And here is another (window -10,2,-300,300) showing the vertical asymptotes at x=-6 and x=-4:




Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Write an equation for a rational function with:
Vertical asymptotes at x+=+-6 and x+=+-4
x intercepts at+x+=+-1 and+x+=+-3
Horizontal asymptote at+y+=+7
a rational function:
f%28x%29=p%28x%29%2Fq%28x%29

given: x intercepts at x+=+-1 and x+=+-3
the x -intercepts exist when the numerator is equal to 0
p%28x%29=%28x-%28-1%29%29%28x-%28-3%29%29
p%28x%29=%28x%2B1%29%28x%2B3%29
p%28x%29=x%5E2+%2B+4+x+%2B+3
To find the vertical asymptote(s) of a rational function, simply set the denominator equal to 0 and solve for x.
you are given vertical asymptotes at x+=+-6 and x+=+-4, so denominator is
q%28x%29=%28x-%28-6%29%29%28x-%28-4%29%29
q%28x%29=%28x%2B6%29%28x%2B4%29
q%28x%29=x%5E2+%2B+10+x+%2B+24

so far f%28x%29=%28x%5E2+%2B+4+x+%2B+3%29%2F%28x%5E2+%2B+10+x+%2B+24%29

you are given horizontal asymptote at y+=+7, and then the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients

so, your function is

f%28x%29=%287x%5E2+%2B+28x+%2B+21%29%2F%28x%5E2+%2B+10+x+%2B+24%29