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

Answer:

Vertical Asymptote:

x=-5 , x=2

Horizontal Asymptote

y=3

Step-by-step explanation:

We are given function as

$f(x)=\frac{3x^2+2x-1}{x^2+3x-10}$

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

$y=\frac{3}{1}$

$y=1$

Answer 2

Answer:

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

Step-by-step 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.