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

tidones0r

tidones0r

Answered question

2022-09-22

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

Answer & Explanation

edytorialkp

edytorialkp

Beginner2022-09-23Added 10 answers

Let's find the holes of this formula. A hole means that the same factor is in the numerator as denominator and they divide out. Such as x 2 x 3 - 4 x or x × x x ( x 2 - 4 ) so there is a hole when x=0.
In our case of 2 x 2 x - 3 , there are no common factors, so there is no hole.
Vertical asymptotes occur when we try to divide a value by 0. So let's see what value of x makes the denominator equal to 0:
x−3=0
x=3
So, there is a vertical asymptote at x=3
Now let's see about the Horizontal asymptote.
I like to use this to help me remember:
BOBO - Bigger on bottom, y=0
BOTN - Bigger on top, none
EATS DC - Exponents are the same, divide coefficients
So in our case, the numerator (top) has a greater exponent (bigger). So there is no Horizontal asmptote (BOTN)
Now let's find the x- intercepts and y- intercepts:
x-intercept is the value of x when y equals 0:
0 = 2 x 2 x - 3
0 = 2 x 2
0 = x 2
x=0
The y-intercept is the value of y when x equals 0
y = 2 ( 0 ) 2 0 - 3
y = 0 - 3
y=0
Now we have all the information we need
To check our answers, let's graph the equation
graph{y=(2x^2)/(x-3)}
We have an x and y intercept at 0, that's right. There's no horizontal asymptote although there is an asymptote for x=3
Our math is correct.

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