Question

Which of the following is true for the rational function shown in the graph below?

Answer 1

@jazzyfa30

Answer 2

Look at the answers with asymptotes. Does either one make sense?

Answer 3

nop i know it has to be either B or D

Answer 4

Ok. Look at B. Can you factor the denominator?

Answer 5

yes

Answer 6

Show me.

Answer 7

That's the numerator.

Answer 8

xD Whoops

Answer 9

Look at the denominator of B.
x^2 + 16
Can this denominator be zero? Is there any value of x that would make this denominator zero?

Answer 10

Now quite:
(x + 4)(x + 4) = (x + 4)^2 = x^2 + 8x + 16
You have x^2 + 16, without the middle term of 8x.

Answer 11

(x - 4)(x - 4) doesn't work either:
(x - 4)(x - 4) = x^2 - 8x + 16
Again, there is amiddle term here that your problem does not have.

Answer 12

2

Answer 13

(x-2)(x-2)

Answer 14

x^2 + 16 for x = 2:
2^2 + 16 =
4 + 16 = 20

Answer 15

Ok. Let's look at B.
The denominator cannot be factored.
x^2 + 16 cannot be factored.
Now think this way: The only way x^2 + 16 could equal zero is if x^2 = -16,
since -16 + 16 = 0. But x^2 cannot equal a negative number. That means, there is no value of x that can make x^2 + 16 equal zero. Since x^2 + 16 is in the denominator, and the denominator cannot be zero, B has no asymptotes.

Answer 16

Now look at D:
The denominator is x^2 - 16.
x^2 - 16 factors into (x + 4)(x - 4)
That means when x = -4 or x = 4, x^2 - 16 equals zero.
That means x = -4 and x = 4 are not allowed.
That means there is an asymptote at x = -4 and an asymptote at x = 4.

Answer 17

ok so d is the answer

Answer 18

Right.