Math / Algebra

QuestionGraph all vertical and horizontal asymptotes of the rational function. $f(x)=\frac{x+1}{-x^{2}-6}$

Studdy Solution

STEP 1

1. The function $f(x) = \frac{x+1}{-x^2-6}$ is a rational function.
2. Vertical asymptotes occur where the denominator is zero and the numerator is not zero.
3. Horizontal asymptotes are determined by the degrees of the polynomial in the numerator and the denominator.

STEP 2

1. Identify potential vertical asymptotes by setting the denominator equal to zero.
2. Determine if any of these points are actual vertical asymptotes by checking the numerator.
3. Determine the horizontal asymptote by comparing the degrees of the numerator and the denominator.

STEP 3

Identify potential vertical asymptotes by setting the denominator equal to zero:
$-x^2 - 6 = 0$

STEP 4

Solve the equation for $x$ :
$-x^2 = 6$ $x^2 = -6$
Since $x^2 = -6$ has no real solutions (as the square of a real number cannot be negative), there are no real values of $x$ that make the denominator zero.

STEP 5

Since there are no real solutions to the equation $-x^2 - 6 = 0$ , there are no vertical asymptotes for this function.

STEP 6

Determine the horizontal asymptote by comparing the degrees of the numerator and the denominator:
- The degree of the numerator $x + 1$ is 1. - The degree of the denominator $-x^2 - 6$ is 2.
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$ .
The graph of the function $f(x) = \frac{x+1}{-x^2-6}$ has a horizontal asymptote at $y = 0$ and no vertical asymptotes.

Was this helpful?

QuestionGraph all vertical and horizontal asymptotes of the rational function. $f(x)=\frac{x+1}{-x^{2}-6}$

Studdy Solution

STEP 1

1. The function $f(x) = \frac{x+1}{-x^2-6}$ is a rational function.
2. Vertical asymptotes occur where the denominator is zero and the numerator is not zero.
3. Horizontal asymptotes are determined by the degrees of the polynomial in the numerator and the denominator.

STEP 2

1. Identify potential vertical asymptotes by setting the denominator equal to zero.
2. Determine if any of these points are actual vertical asymptotes by checking the numerator.
3. Determine the horizontal asymptote by comparing the degrees of the numerator and the denominator.

STEP 3

Identify potential vertical asymptotes by setting the denominator equal to zero:
$-x^2 - 6 = 0$

STEP 4

Solve the equation for $x$ :
$-x^2 = 6$ $x^2 = -6$
Since $x^2 = -6$ has no real solutions (as the square of a real number cannot be negative), there are no real values of $x$ that make the denominator zero.

STEP 5

Since there are no real solutions to the equation $-x^2 - 6 = 0$ , there are no vertical asymptotes for this function.

STEP 6

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionGraph all vertical and horizontal asymptotes of the rational function. f(x)=x+1−x2−6f(x)=\frac{x+1}{-x^{2}-6}f(x)=−x2−6x+1​

Studdy Solution

STEP 1

STEP 2

1. Identify potential vertical asymptotes by setting the denominator equal to zero.2. Determine if any of these points are actual vertical asymptotes by checking the numerator.3. Determine the horizontal asymptote by comparing the degrees of the numerator and the denominator.

STEP 3

Identify potential vertical asymptotes by setting the denominator equal to zero:−x2−6=0 -x^2 - 6 = 0 −x2−6=0

STEP 4

Solve the equation for x x x:−x2=6 -x^2 = 6 −x2=6 x2=−6 x^2 = -6 x2=−6Since x2=−6 x^2 = -6 x2=−6 has no real solutions (as the square of a real number cannot be negative), there are no real values of x x x that make the denominator zero.

STEP 5

Since there are no real solutions to the equation −x2−6=0 -x^2 - 6 = 0 −x2−6=0, there are no vertical asymptotes for this function.

STEP 6

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionGraph all vertical and horizontal asymptotes of the rational function. $f(x)=\frac{x+1}{-x^{2}-6}$

1. Identify potential vertical asymptotes by setting the denominator equal to zero.
2. Determine if any of these points are actual vertical asymptotes by checking the numerator.
3. Determine the horizontal asymptote by comparing the degrees of the numerator and the denominator.

Identify potential vertical asymptotes by setting the denominator equal to zero:
$-x^2 - 6 = 0$

Solve the equation for $x$ :
$-x^2 = 6$ $x^2 = -6$
Since $x^2 = -6$ has no real solutions (as the square of a real number cannot be negative), there are no real values of $x$ that make the denominator zero.

Since there are no real solutions to the equation $-x^2 - 6 = 0$ , there are no vertical asymptotes for this function.