How do you graph f(x)=(3x^2−2x)/(x−1) using holes, vertical and horizontal asymptotes, x and y intercepts?

bamakhosimz

bamakhosimz

Answered question

2022-09-12

How do you graph f ( x ) = 3 x 2 - 2 x x - 1 using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer & Explanation

Zayden Dorsey

Zayden Dorsey

Beginner2022-09-13Added 18 answers

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
solve x - 1 = 0 x = 1 is the asymptote
Since the degree of the numerator > degree of the denominator there will be an oblique asymptote but no horizontal asymptote.
To obtain oblique asymptote, divide numerator by denominator.
Consider the numerator
3 x ( x - 1 ) + 3 x - 2 x
= 3 x ( x - 1 ) + 1 ( x - 1 ) + 1
oblique asymptote is y = 3 x + 1
Intercepts
x = 0 y = 0 ( 0 , 0 ) y-intercept
y = 0 3 x 2 - 2 x = 0 x ( 3 x - 2 ) = 0
x = 0 , x = 2 3 x-intercepts
graph{(3x^2-2x)/(x-1) [-20, 20, -10, 10]}

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