Question

Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. $$ R(x)=\frac{8 x^{2}+26 x-7}{4 x-1} $$

Short answer

Vertical asymptote: \(x = \frac{1}{4}\)No horizontal asymptoteOblique asymptote: \(y = 2x + 7\)

Step-by-step solution

- Find the vertical asymptotes

Vertical asymptotes occur where the denominator is zero. Set the denominator equal to zero and solve for x: \[4x - 1 = 0\] Solving for x: \[x = \frac{1}{4}\] Therefore, there is a vertical asymptote at \[x = \frac{1}{4}\].

- Determine the horizontal asymptote

Horizontal asymptotes are determined by comparing the degrees of the numerator and denominator. In this function, the numerator is \(8x^2 + 26x - 7\), which is of degree 2, and the denominator is \(4x - 1\), which is of degree 1. Since the degree of the numerator (2) is higher than the degree of the denominator (1), there is no horizontal asymptote.

- Identify the oblique asymptote

Oblique (or slant) asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator. Here, the numerator is of degree 2 and the denominator is of degree 1, so we need to perform polynomial long division:Performing the division of \(8x^2 + 26x - 7\) by \(4x - 1\):1. Divide: \(8x^2 ÷ 4x = 2x\)2. Multiply: \(2x * (4x - 1) = 8x^2 - 2x\)3. Subtract: \((8x^2 + 26x - 7) - (8x^2 - 2x) = 28x - 7\)4. Divide: \(28x ÷ 4x = 7\)5. Multiply: \(7 * (4x - 1) = 28x - 7\)6. Subtract: \((28x - 7) - (28x - 7) = 0\)The quotient is \(2x + 7\), hence the oblique asymptote is the line \(y = 2x + 7\).

Key concepts

Vertical Asymptotes

Vertical asymptotes occur where the denominator of a rational function equals zero. In simpler terms, they are the values of x that make the denominator zero and the function undefined. For the function given, \(R(x)=\frac{8x^2 + 26x - 7}{4x - 1}\), we find vertical asymptotes by setting the denominator, \(4x - 1\), equal to zero:\rewline\r\[4x - 1 = 0\] \rSolving for x, we get: \rewline\r\[x = \frac{1}{4}\] \rSo, there is a vertical asymptote at \(x = \frac{1}{4}\).\rewline\r In general, vertical asymptotes help to understand the behavior of graphs and where they shoot to infinity or negative infinity.

Horizontal Asymptotes

Horizontal asymptotes are lines that the graph of a function approaches as x tends to ±∞. They reflect the end behavior of a rational function. To determine if a horizontal asymptote exists, compare the degrees (highest power terms) of the numerator and the denominator. For \(R(x)=\frac{8 x^{2}+26 x-7}{4 x-1}\):\rewline\r

\r
• Numerator degree: 2 (since the highest power of x is x^2)
\r
• Denominator degree: 1 (since the highest power of x is x^1)
\r

\r \r When the degree of the numerator is greater than the degree of the denominator (as it is here), there is no horizontal asymptote. The graph doesn't settle into a horizontal line as x moves to infinity or negative infinity.

Oblique Asymptotes

Oblique asymptotes, also known as slant asymptotes, occur when the degree of the numerator of a rational function is exactly one more than the degree of the denominator. For \(R(x)=\frac{8 x^{2}+26 x-7}{4 x-1}\), since the degree of the numerator (2) is one more than the degree of the denominator (1), we expect an oblique asymptote. We find it using polynomial long division:\r\r1. Divide: \(8x^2 ÷ 4x = 2x\)\rewline\r2. Multiply: \(2x * (4x - 1) = 8x^2 - 2x\)\rewline\r3. Subtract: \(8x^2 + 26x - 7 - (8x^2 - 2x) = 28x - 7\)\rewline\r4. Divide: \(28x ÷ 4x = 7\)\rewline\r5. Multiply: \(7 * (4x - 1) = 28x - 7\)\rewline\r6. Subtract: \(28x - 7 - (28x - 7) = 0\)\r\rThe quotient is \(2x + 7\). Hence, the equation of the oblique asymptote is \(y = 2x + 7\). This line describes the graph's slant direction as x approaches infinity or negative infinity.

Polynomial Long Division

Polynomial long division is a method to divide one polynomial by another, similar to long division with numbers. It helps to identify oblique asymptotes and simplify expressions. Here’s a step-by-step explanation for dividing \(8x^2 + 26x - 7\) by \(4x - 1 \):\rewline\r

\r
• First, divide the leading term of the numerator (8x^2) by the leading term of the denominator (4x), which results in 2x.\r
• Then, multiply the entire divisor (4x - 1) by this result (2x), giving us 8x^2 - 2x.\r
• Subtract this from the original numerator polynomial. \(8x^2 + 26x - 7 - (8x^2 - 2x) = 28x - 7\).
\r
• Repeat the process with the new polynomial (28x - 7): Divide 28x by 4x, yielding 7. Multiply (4x - 1) by 7, giving 28x - 7.\r
• Subtracting again, (28x - 7) - (28x - 7) results in 0, showing the division is complete.\r

\rThis yields the quotient 2x + 7, thus indicating the oblique asymptote as y = 2x + 7.\rewline\rThis method is useful not just for finding asymptotes but also for simplifying polynomial expressions in algebra.