Question

Which statement make correctly uses limits determine a vertical asymptote of G(x)=-7(x-5)^2(x+6)/(x-5)(x+5)

Answer 1

Step-by-step explanation: A

Answer 2

There is a vertical Asymptote at x = 5 because
`\lim_(x \to 5^(-) ) G(x) = \infty\\lim_(x \to 5^(+) ) G(x) = -\infty`

There is a vertical Asymptote at x = -5 because
`\lim_(x \to -5^(-) ) G(x) = \infty\\\lim_(x \to -5^(+) ) G(x) = -\infty`

Explanation:

The exact question is as follows :

Given - G(x) = -7(x-5)^2(x+6)/(x-5)(x+5)

To find - Which statement make correctly uses limits determine a vertical Asymptote of G(x)

Solution -

Vertical Asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function.

Here the given function is the rational function

Denominator of G(x) = (x - 5)(x + 5)

So,

Put denominator = 0, we get

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

⇒x = 5, -5

∴ we get

G(x) has vertical Asymptotes at x = 5 and x = -5

Now,

At x= 5

`\lim_(x \to 5^(-) ) G(x) = \infty\\\lim_(x \to 5^(+) ) G(x) = -\infty`

And

At x = -5

`\lim_(x \to -5^(-) ) G(x) = \infty\\\lim_(x \to -5^(+) ) G(x) = -\infty`

∴ we get

There is a vertical Asymptote at x = 5 because
`\lim_(x \to 5^(-) ) G(x) = \infty\\lim_(x \to 5^(+) ) G(x) = -\infty`

There is a vertical Asymptote at x = -5 because
`\lim_(x \to -5^(-) ) G(x) = \infty\\\lim_(x \to -5^(+) ) G(x) = -\infty`