anonymous
  • anonymous
Create a function that has a graph with features that include -vertical retricemptote at y axis x=3 -horizontal assymptote at y=2 -x intercept of (-1/2, 0) & (1,0) QUESTION IS REGAURDING RATIONAL FUNCTIONS/QUADRATIC FUNCTIONS ...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
asymptotes
anonymous
  • anonymous
lol
anonymous
  • anonymous
vertical at x = 3 means denominator should have a factor of \[x-3\] in it

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
horizontal asymptote at \[y=2\] means the degree of the numerator and denominator must be the same, and the ratio of leading coefficients has to be 2
anonymous
  • anonymous
so maybe \[f(x)=\frac{2x}{x-3}\] would work
anonymous
  • anonymous
unfortunately it does not, because \[f(1)=\frac{2}{-2}=-1\] and you want \[f(1)=0\] so maybe try \[f(x)=\frac{2x-2}{x-3}\]
anonymous
  • anonymous
now at least \[f(1)=\frac{2-2}{1-3}=0\] how about \[f(-\frac{1}{2})\]?
anonymous
  • anonymous
nope that doesn't work, i am a moron numerator should be \[(2x+1)(x-1)\]
anonymous
  • anonymous
that will give you the correct zeros,
anonymous
  • anonymous
so now maybe \[f(x)=\frac{(2x+1)(x-1)}{(x-3)^2}\]
anonymous
  • anonymous
that one will have the correct zeros, and also the correct horizontal and vertical asymptote. i was forgetting that if you have two zeros you need a polynomial of degree 2 !

Looking for something else?

Not the answer you are looking for? Search for more explanations.