Write the equation of a function with the following requirements: A domain of all real numbers over 3, a range of all real numbers, an x-intercept of (5,0), and a vertical asymptote at x = 3.
(x-5) will cross x axis at 5
1/(x-3) goes nuts at 3
so try
f(x) = (x-5)/(x-3)
check domain and range
any x except x = 3 so domain ok
now sketch a graph to see range
if x is big + or -, f(x) = 1
if x is little less than 3, f(x)--> -oo
if x is a little more than 3, f(x)->+oo
and is 0 at x = 5
so range is all real numbers, ok
if x is a little less than
How about
vertical asymptote at x=3, domain is x>3, x-intercept of (4,0)
y = log(x-3)
to move the intercept to (5,0), we need to double the distance from the asymptote to the intercept.
y = log((x-3)/2)
see
http://www.wolframalpha.com/input/?i=log(1%2F2+(x-3))+for+x%3E3
To create a function with the given requirements, we can start by considering the x-intercept of (5,0). For a function to have an x-intercept at (5,0), the function must pass through the point (5,0). This means that when x = 5, the value of the function, denoted as f(x), should be equal to 0.
Now let's ensure we have a vertical asymptote at x = 3. A vertical asymptote occurs when the function approaches a certain x-value but does not cross it. In our case, the vertical asymptote is x = 3. To represent this vertical asymptote, we can include a factor in the function that results in a denominator of (x - 3).
Based on the requirements, we need to create a rational function (fraction of two polynomials) where the numerator and denominator are such that the x-intercept is (5,0) and the vertical asymptote is x = 3.
Let's proceed step by step:
1. Start with the denominator: (x - 3). This creates the desired vertical asymptote at x = 3.
2. For the numerator, we are free to choose any polynomial that meets the requirement of having the x-intercept at (5,0). Since the range is all real numbers, we can use a constant multiple of (x - 5) for simplicity. Let's multiply the numerator by -1 to ensure that the x-intercept is (5,0).
Putting it all together, we have:
f(x) = (-1(x - 5))/ (x - 3)
This equation meets all the given requirements:
- The domain is all real numbers over 3 since there are no restrictions on x besides the vertical asymptote.
- The range is all real numbers because the numerator and denominator are both polynomials of degree 1.
- The x-intercept is (5,0) since f(5) = 0.
- The vertical asymptote is x = 3 because the denominator is (x - 3).
I hope this explanation helps! Let me know if you have any further questions.