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11Alexandr11 [23.1K]
3 years ago
5

Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2. ​

Mathematics
1 answer:
insens350 [35]3 years ago
6 0

Answer:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

Step-by-step explanation:

Given

p(x) = \frac{x^2-2x-3}{x+2} -- Missing from the question

Required

The behavior of the function around its vertical asymptote at x = -2

p(x) = \frac{x^2-2x-3}{x+2}

Expand the numerator

p(x) = \frac{x^2 + x -3x - 3}{x+2}

Factorize

p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}

Factor out x + 1

p(x) = \frac{(x -3)(x + 1)}{x+2}

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

----------------------------------------------------------------------------------------------------------

As x approaches -2 implies that:

x -> -2^{-} Say x = -3

p(x) = \frac{(x -3)(x + 1)}{x+2}

p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

x->-2^{-}, p(x)->-\infty

Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: x>-2

Say x = -2.1

p(-2.1) = \frac{(-2.1-3)(-2.1+1)}{-2.1+2} = \frac{-5.1 * -1.1}{-0.1} = \frac{+5.61}{-0.1} = -56.1

We have a negative value (-56.1); This will be called negative infinity

This implies that as x leaves -2, p(x) approaches negative infinity

x->-2^{+}, p(x)->-\infty

So, the behavior is:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

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