Question

How do I identify the intercepts, removable continuities, and asymptotes?
(2x^3-3x+1)/x^3-5x+7

Answer 1

@ganeshie8

Answer 2

are you allowed to graph the function ?

Answer 3

nope :/ i think i'm supposed to solve it based on just the equation

Answer 4

Finding horizontal asymptotes is easy for a rational function

Answer 5

\[y = \dfrac{\color{Red}{2}x^3-3x+1}{\color{red}{1}x^3-5x+7}\]

As \(x\) gets large, the lower degree terms don't matter much, so the value approaches

\[y \approx \dfrac{\color{Red}{2}x^3}{\color{red}{1}x^3} = \dfrac{\color{Red}{2}}{\color{red}{1}}=\color{red}{2}\]

so \(\color{red}{y=2}\) is an horizontal asymptote

Answer 6

thank you!! is the vertical asymptote about 2.75?

Answer 7

x^3-5x=-7 right? and solve for x

Answer 8

it should be around x= -2.75

but how did u get that ?

Answer 9

Yes

Answer 10

okay thanks!! i'm super confused on finding the x intercepts though

Answer 11

did you find y intercepts ?

Answer 12

yeah i got 1/7 by plugging in 0 in place of x

Answer 13

Good. use the same trick
simply plugin y=0 to get the x intercepts

Answer 14

\[y = \dfrac{\color{black}{2}x^3-3x+1}{\color{black}{1}x^3-5x+7}\]

plugin y=0 for x intercepts

\[0 = \dfrac{\color{black}{2}x^3-3x+1}{\color{black}{1}x^3-5x+7}\]

Answer 15

which is same as

\[0 =\color{black}{2}x^3-3x+1\]

solve

Answer 16

wolfram gives 3 x intercepts
http://www.wolframalpha.com/input/?i=solve+2x%5E3-3x%2B1%3D0

Answer 17

could i use the formula bc of the plus and minus?

Answer 18

quadratic formula*

Answer 19

nope, the equation is cubic, not quadratic.

i would simply use calculator/wolfram

Answer 20

i'm confused on how i would get 3 answers :/

Answer 21

why ?
it is fine to have multiple x intercepts

Answer 22

that just means, the graph cuts the x axis at 3 different places

Answer 23

i used mathway and it gave me (1,0) and (-1+-sqrt3)/2 but i'm not sure how it got that

Answer 24

oh i see it haha thank you so much :)