How do you graph f(x)=(x^2+4x+3)/(−3x−6) using holes, vertical and horizontal asymptotes, x and y intercepts?

dripcima24

dripcima24

Answered question

2022-09-29

How do you graph f ( x ) = x 2 + 4 x + 3 - 3 x - 6 using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer & Explanation

Nolan Tyler

Nolan Tyler

Beginner2022-09-30Added 9 answers

f ( x ) = x 2 + 4 x + 3 - 3 x - 6
f ( x ) = - 1 3 ( x 2 + 4 x + 3 x + 2 )
f ( x ) = - 1 3 ( x + 2 ) ( x + 2 ) - 1 x + 2
f ( x ) = - 1 3 ( x + 2 - 1 x + 2 )
f ( x ) = - 1 3 ( x + 2 ) + 1 3 ( x + 2 )

For oblique asymptote, it is y = - 1 3 ( x + 2 ) which can be found by letting x .
For vertical asymptote, the denominator cannot equal to 0. so x + 2 0 so at x=−2, there is an asymptote.

For intercepts,
When x=0, y = - 1 2
When y=0, x 2 + 4 x + 3 = 0
(x+3)(x+1)=0 so x=−1 and x=−3

Plotting your x and y intercepts as well as your asymptotes, you should have a general idea as to how your graph should appear. The ends of your graph should be approaching your asymptotes but they should NEVER TOUCH the asymptotes.

Below is the graph
graph{(x^2+4x+3)/(-3x-6) [-10, 10, -5, 5]}

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