Chapter 3: Problem 26
Find all horizontal and vertical asymptotes (if any). $$s(x)=\frac{8 x^{2}+1}{4 x^{2}+2 x-6}$$
Short Answer
Expert verified
Horizontal asymptote: \(y = 2\). Vertical asymptotes: \(x = 1\) and \(x = -\frac{3}{2}\).
Step by step solution
01
Identify Horizontal Asymptotes
To find the horizontal asymptote, compare the degrees of the numerator and the denominator. In the function \(s(x)=\frac{8x^2+1}{4x^2+2x-6}\), both the numerator and denominator are quadratic (degree 2). The horizontal asymptote is found by taking the coefficients of the highest degree terms: \(y = \frac{8}{4} = 2\).
02
Identify Vertical Asymptotes
Vertical asymptotes occur where the denominator equals zero, provided the numerator is non-zero at those values. Solve \(4x^2+2x-6=0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) with \(a=4, b=2, c=-6\).
03
Calculate Discriminant
Calculate the discriminant of the quadratic formula: \(b^2 - 4ac = 2^2 - 4(4)(-6) = 4 + 96 = 100\). Since the discriminant is positive, there are two real roots.
04
Solve for Vertical Asymptotes
Substitute into the quadratic formula: \(x = \frac{-2 \pm \sqrt{100}}{8}\). Simplify to get \(x = \frac{-2 \pm 10}{8}\), resulting in \(x = 1\) and \(x = -\frac{3}{2}\). These are the vertical asymptotes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Horizontal Asymptotes
Horizontal asymptotes give us an idea of how a function behaves as the input value, represented by the variable usually denoted as \( x \), approaches positive infinity \( (+\infty) \) or negative infinity \( (-\infty) \). In the context of rational functions such as \( s(x) = \frac{8x^2+1}{4x^2+2x-6} \), the horizontal asymptote is determined by comparing the degrees of the numerator and the denominator.
For the function \( s(x) \), both the numerator \( 8x^2 + 1 \) and the denominator \( 4x^2 + 2x - 6 \) are quadratic, meaning they both have a degree of 2. In such cases, the horizontal asymptote is identified by dividing the leading coefficients of the numerator and the denominator.
This indicates that as \( x \) becomes very large or very small, the output \( s(x) \) approaches the line \( y = 2 \).
For the function \( s(x) \), both the numerator \( 8x^2 + 1 \) and the denominator \( 4x^2 + 2x - 6 \) are quadratic, meaning they both have a degree of 2. In such cases, the horizontal asymptote is identified by dividing the leading coefficients of the numerator and the denominator.
- The leading coefficient of the numerator is 8.
- The leading coefficient of the denominator is 4.
This indicates that as \( x \) becomes very large or very small, the output \( s(x) \) approaches the line \( y = 2 \).
Vertical Asymptotes
Vertical asymptotes are locations on the graph where the function approaches infinity or negative infinity, causing a break in the graph. They occur at roots of the denominator where the numerator is non-zero, making the expression undefined due to division by zero.
In the function \( s(x) = \frac{8x^2+1}{4x^2+2x-6} \), to find vertical asymptotes, solve the equation \( 4x^2 + 2x - 6 = 0 \).
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 4 \), \( b = 2 \), \( c = -6 \), you first compute the discriminant:
\[ b^2 - 4ac = 2^2 - 4(4)(-6) = 4 + 96 = 100 \]
With a positive discriminant, the function has two real roots, thus two vertical asymptotes.
Solving gives:
\[ x = \frac{-2 \pm \sqrt{100}}{8} \]
This simplifies to \( x = 1 \) and \( x = -\frac{3}{2} \).
These values represent the vertical asymptotes of the function. The graph of \( s(x) \) shoots to infinity or negative infinity as \( x \) approaches these values.
In the function \( s(x) = \frac{8x^2+1}{4x^2+2x-6} \), to find vertical asymptotes, solve the equation \( 4x^2 + 2x - 6 = 0 \).
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 4 \), \( b = 2 \), \( c = -6 \), you first compute the discriminant:
\[ b^2 - 4ac = 2^2 - 4(4)(-6) = 4 + 96 = 100 \]
With a positive discriminant, the function has two real roots, thus two vertical asymptotes.
Solving gives:
\[ x = \frac{-2 \pm \sqrt{100}}{8} \]
This simplifies to \( x = 1 \) and \( x = -\frac{3}{2} \).
These values represent the vertical asymptotes of the function. The graph of \( s(x) \) shoots to infinity or negative infinity as \( x \) approaches these values.
Quadratic Functions
A quadratic function is a type of polynomial where the highest power of the variable \( x \) is 2. Such functions take the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).
The function \( s(x) = \frac{8x^2+1}{4x^2+2x-6} \) is built on two quadratic expressions, one serving as the numerator and the other as the denominator of a rational function.
Key characteristics of quadratic functions include:
The function \( s(x) = \frac{8x^2+1}{4x^2+2x-6} \) is built on two quadratic expressions, one serving as the numerator and the other as the denominator of a rational function.
Key characteristics of quadratic functions include:
- The graph of a quadratic function is a parabola.
- The direction of the parabola (opening upwards or downwards) depends on the sign of the coefficient \( a \). Positive \( a \) results in a parabola opening upwards, while negative \( a \) results in one that opens downwards.
- The roots of the quadratic function can be found using methods like factoring, completing the square, or using the quadratic formula.