How do you identify all asymptotes or holes for #f(x)=(x^2+3x-4)/(2x^2+10x+8)#?

Answer 1

vertical asymptote at x = - 1
horizontal asymptote at #y=1/2#
hole at x = -4

The first step is to factorise the numerator/denominator of f(x).

#f(x)=(x^2+3x-4)/(2x^2+10x+8)=(cancel((x+4))(x-1))/(2cancel((x+4))(x+1))=(x-1)/(2x+2)#

excluded value is x ≠ -4. This means that the original function has a hole at x = -4, while the simplified version does not.

The denominator of simplified f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : #2x+2=0rArrx=-1" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by x

#f(x)=(x/x-1/x)/((2x)/x+2/x)=(1-1/x)/(2+2/x)#
as #xto+-oo,f(x)to(1-0)/(2+0)#
#rArry=1/2" is the asymptote"# graph{(x-1)/(2x+2) [-10, 10, -5, 5]}
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Answer 2

To identify all asymptotes or holes for (f(x) = \frac{x^2 + 3x - 4}{2x^2 + 10x + 8}), follow these steps:

  1. Identify Vertical Asymptotes: Vertical asymptotes occur where the denominator of the function equals zero but the numerator does not. Solve (2x^2 + 10x + 8 = 0) to find vertical asymptotes.

  2. Identify Horizontal Asymptotes: Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. Determine the horizontal asymptote by comparing the degrees of the numerator and denominator.

  3. Identify Holes: Holes occur when factors cancel out in both the numerator and denominator of the function, leaving a removable discontinuity. Cancel out common factors and determine the value of (x) that makes the function undefined. This will give you the (x)-coordinate of the hole. To find the (y)-coordinate of the hole, evaluate the function at that (x)-value.

Let's apply these steps to (f(x)):

  1. Vertical Asymptotes: Solve (2x^2 + 10x + 8 = 0) to find vertical asymptotes, if any.

[2x^2 + 10x + 8 = 0]

[x^2 + 5x + 4 = 0]

((x + 1)(x + 4) = 0)

So, vertical asymptotes occur at (x = -1) and (x = -4).

  1. Horizontal Asymptotes: Compare the degrees of the numerator and denominator.

Degree of the numerator: 2 Degree of the denominator: 2

Since the degrees are the same, the horizontal asymptote is determined by the ratio of the leading coefficients.

The horizontal asymptote is (y = \frac{1}{2}).

  1. Holes: Factor both the numerator and denominator and check if any common factors cancel out.

[f(x) = \frac{x^2 + 3x - 4}{2x^2 + 10x + 8}]

[f(x) = \frac{(x + 4)(x - 1)}{2(x + 1)(x + 4)}]

We see that there's a common factor of (x + 4) in both the numerator and denominator, which cancels out.

So, there's a hole at (x = -4).

To find the (y)-coordinate of the hole, plug (x = -4) into the simplified function:

[f(-4) = \frac{(-4 + 4)(-4 - 1)}{2(-4 + 1)}]

[f(-4) = \frac{0(-5)}{2(-3)}]

[f(-4) = \frac{0}{-6}]

[f(-4) = 0]

So, there's a hole at ((-4, 0)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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