Question

What are the equations of the asymptotes of y=−1.6x+0.3−3.4?

Which of the following hyperbolas has a center at (−2,3)?

Answer 1

Answer: 1) D. x = -0.3 and y = -3.4

2) B. $y=\frac{4}{x+2}+3$

Step-by-step explanation:

1) Given equation,

$y=\frac{-1.6}{x+0.3}-3.4$

For vertical asymptote,

$y\rightarrow \infty$

It is possible for the given function, if the denominator = 0

⇒ x + 0.3 = 0

⇒ x = -0.3

Now, for horizontal asymptote,

$x\rightarrow \infty$

By substituting this in the given equation of the function,

We get,

$y=\frac{-1.6}{\infty+0.3}-3.4$

$y=\frac{-1.6}{\infty}-3.4$

$y=0-3.4$

$y=-3.4$

Hence, the vertical and horizontal asymptote of the given function are x = -3.0 and y = -3.4 respectively.

⇒ Option D is correct.

2) If a function is defined as,

(x-h)(y-k) = a

Then (h,k) be the center of this function,

In option A,

Given function is,

$y = \frac{4}{x+3}+2$

$\implies y-2 = \frac{4}{x+3}$

$\implies (x+3)(y-2) = 4$

⇒ The center of the function is (-3,2)

In option B,

Given function is,

$y = \frac{4}{x+2}+3$

$\implies y-3 = \frac{4}{x+2}$

$\implies (x+2)(y-3) = 4$

⇒ The center of the function is (-2,3)

In option C,

Given function is,

$y = \frac{4}{x-2}+3$

$\implies y-3 = \frac{4}{x-2}$

$\implies (x-2)(y-3) = 4$

⇒ The center of the function is (2,3)

In option D,

Given function is,

$y = \frac{4}{x+3}-2$

$\implies y+2 = \frac{4}{x+3}$

$\implies (x+3)(y+2) = 4$

⇒ The center of the function is (-3,-2)

⇒ Thus, by the above explanation the function in option B has the center of (-2,3)