How do you graph #f(x)=(2x-3)/(x^2+x-6)# using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

Below

#f(x)=(2x-3)/(x^2+x-6)#
For vertical asymptotes, #x^2+x-6!=0# as the denominator cannot equal to 0. If it equals to 0, then the function at the point is undefined. To find the vertical asymptotes, we find at what points the function is undefined
#x^2+x-6=0# #(x+3)(x-2)=0# #x=-3# and #x=2#
For horizontal asymptotes, #y=0# is the only asymptote.
Think of it this way: When you sub random numbers into #x#, the denominator will always be bigger than the numerator. Hence, when you divide a small number by a larger number, then the answer will tend to 0 (be close to 0)
For intercepts, When #y=0#, #2x-3=0#, #x=3/2# When #x=0#, #y=1/2#

Plotting your intercepts and drawing in your asymptotes, you should get something like below. Remember, your asymptotes only affect/influence the endpoints of your graph and not anywhere else on your graph

graph{(2x-3)/(x^2+x-6) [-10, 10, -5, 5]}

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To graph the function f(x) = (2x-3)/(x^2+x-6), we can follow these steps:

  1. Factorize the denominator (x^2+x-6) to find any potential holes or vertical asymptotes: x^2 + x - 6 = (x-2)(x+3) Therefore, the function has potential holes at x = 2 and x = -3.

  2. Determine the vertical asymptotes by finding the values of x that make the denominator equal to zero: x^2 + x - 6 = 0 Solving this quadratic equation, we find x = 2 and x = -3. Hence, the vertical asymptotes are x = 2 and x = -3.

  3. Find the x-intercepts by setting the numerator equal to zero and solving for x: 2x - 3 = 0 Solving this equation, we get x = 3/2. Therefore, the x-intercept is (3/2, 0).

  4. Find the y-intercept by evaluating the function at x = 0: f(0) = (2(0) - 3)/(0^2 + 0 - 6) = -3/6 = -1/2. Hence, the y-intercept is (0, -1/2).

  5. Determine the behavior of the function as x approaches positive or negative infinity to find the horizontal asymptotes: As x approaches positive or negative infinity, the function approaches zero. Therefore, the horizontal asymptote is y = 0.

  6. Plot the obtained points: holes at x = 2 and x = -3, vertical asymptotes at x = 2 and x = -3, x-intercept at (3/2, 0), y-intercept at (0, -1/2), and the horizontal asymptote at y = 0.

By following these steps, you can graph the function f(x) = (2x-3)/(x^2+x-6) accurately.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7