Question

What are the equations of the asymptotes of the graph of the function f(x)=3x^2-2x-1/x^2+3x-10 ? x = –5, x = 2 and y = 3 x = –2, x = 5 and y = 3 x = 3, y = –5, and y = 2 x = 3, y = –2, and y = 5

Answer 1

Vertical Asymptote:

x=-5 , x=2

Horizontal Asymptote

y=3

Explanation:

We are given function as

`f(x)=(3x^2+2x-1)/(x^2+3x-10)`

Vertical asymptote:

For finding vertical asymptote , we can set denominator =0

and then we can solve for x

`x^2+3x-10=0`

we can factor it

`(x+5)(x-2)=0`

`x=-5,x=2`

Horizontal asymptote:

we can see that

degree of numerator =2

degree of denominator =2

So, degree of both numerator and denominator are same

So, for finding horizontal asymptote , we can find ratio of leading coefficient

Leading coefficient of numerator=3

leading coefficient of denominator =1

so, horizontal asymptote will be

`y=(3)/(1)`

`y=1`

Answer 2

x = -5, x = 2 and y = 3

Explanation:

The given function f(x) = (3x^2 - 2x - 1) /(x^2 + 3x - 10)

First let's find the horizontal asymptote.

Here the degree of the numerator and the denominator are the same. Which is 2.

Therefore, the horizontal asymptote = The leading coefficient of the numerator ÷ The leading coefficient of the denominator.

y = 3/1

y = 3

Now let's find the vertical asymptotes.

To find the vertical asymptotes set the denominator equal to zero and find the values of x.

x^2 + 3x - 10 = 0

(x - 2)(x + 5) = 0

x - 2 = 0 and x + 5 = 0

Solving the above, we get

x = 2 and x = -5

Therefore, the vertical asymptotes are at x =2 and x = -5

Answer: x = -5, x = 2 and y = 3.

Thank you.