Find the horizontal asymptote of each function. f(x)=8x2+12x2−x−6f\left(x\right)=\frac{8x^2+1}{2x^2-x-6}f(x)=2x2−x−68x2+1

Horizontal Asymptote at y=4y=4y=4

Find the horizontal asymptote of each function. f ( x ) = 8 x 2 + 1 2 x 2 − x − 6 f\left(x\right)=\frac{8x^2+1}{2x^2-x-6} f ( x ) = 2 x 2 − x − 6 8 x 2 + 1

In this problem, we're asked to find the horizontal asymptote of the function f of x is equal to 8x squared plus 1 over 2x squared minus x minus 6. So looking at the degree of the numerator and the denominator here, the degree of my numerator is 2, and the degree of my denominator is also 2. So that tells me that these degrees are equal to each other, which I now need to divide my leading coefficients here. So taking a look at my leading coefficients, I have 8 and 2. So my leading coefficient of my numerator divided by the leading coefficient of my denominator, 8 over 2, will give me a horizontal asymptote at y equals 4. I'll see you in the next video.

Find the horizontal asymptote of each function. f(x)=8x2+12x2−x−6f\left(x\right)=\frac{$$8x^2+1$$}{$$2x^2-x-6$$}f(x)=2x2−x−68x2+1

Horizontal Asymptote at y=4y=4y=4

Horizontal Asymptote at y=0y=0y=0

Horizontal Asymptote at y=14y=\frac14y=41

5. Rational Functions

Asymptotes

College Algebra

5. Rational Functions

Asymptotes

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Multiple Choice

Find the horizontal asymptote of each function.
$f$\left$(x$\right$)=$\frac{8x^2+1}{2x^2-x-6}$$

Horizontal Asymptote at $y=0$

Horizontal Asymptote at $y=$\frac$14$

Horizontal Asymptote at $y=4$

Verified step by step guidance

Identify the degrees of the numerator and the denominator. The degree of a polynomial is the highest power of the variable in the polynomial. For the function f(x) = $\frac{8x^2 + 1}{2x^2 - x - 6}$, both the numerator and the denominator are quadratic polynomials, meaning their degree is 2.

Compare the degrees of the numerator and the denominator. If the degrees are equal, the horizontal asymptote is determined by the ratio of the leading coefficients. The leading coefficient is the coefficient of the term with the highest degree.

For the numerator 8x^2 + 1, the leading coefficient is 8. For the denominator 2x^2 - x - 6, the leading coefficient is 2.

Calculate the horizontal asymptote by dividing the leading coefficient of the numerator by the leading coefficient of the denominator: $\frac{8}{2}$ = 4.

Conclude that the horizontal asymptote of the function f(x) = $\frac{8x^2 + 1}{2x^2 - x - 6}$ is y = 4.

6:24m

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Struggling with College Algebra?

Find the horizontal asymptote of each function. f(x)=8x2+12x2−x−6f\(\left\)(x\(\right\))=\(\frac{8x^2+1}{2x^2-x-6}\)f(x)=2x2−x−68x2+1

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Asymptotes practice set

Find the horizontal asymptote of each function.
$f\(\left\)(x\(\right\))=\(\frac{8x^2+1}{2x^2-x-6}\)$