Problem 17
Question
Find the domain and the vertical and horizontal asymptotes (if any). $$f(x)=\frac{2 x+7}{2 x^{2}+5 x-3}$$
Step-by-Step Solution
Verified Answer
The domain of the function is all real numbers except \(x = 0.5\) and \(x = -3\). The vertical asymptotes of the function are \(x = 0.5\) and \(x = -3\). The horizontal asymptote of the function is \(y = 0\).
1Step 1: Finding the Domain
To find the domain of the function \( f(x) = \frac{2x + 7}{2x^{2} + 5x - 3} \), we must set the denominator equal to zero and solve for 'x'. This gives us the values that 'x' cannot be, as they would make the function undefined. So, let's solve for 'x' in the equation \(2x^{2} + 5x - 3 = 0\). Using quadratic formula \( x = \frac{-b ± \sqrt{b^{2} - 4ac}}{2a}\), we get \( x = \frac{-5 ± \sqrt{(5)^{2} - 4(2)(-3)}}{2(2)}\), which simplifies to \(x = (-5 + 7)/4\) or \(x = (-5 - 7)/4\), giving us \(x = 0.5\) and \(x = -3\). Hence, the domain is all real numbers except \(x = 0.5\) and \(x = -3\).
2Step 2: Finding Vertical Asymptotes
As mentioned above, the vertical asymptotes occur at the 'x' values where the function is undefined. We determined these values already in step 1, and they are \(x = 0.5\) and \(x = -3\). So, these are our vertical asymptotes.
3Step 3: Finding Horizontal Asymptotes
To find the horizontal asymptotes, we need to look at the behavior of the function as 'x' approaches infinity. If the degree of the numerator is less than the degree of the denominator, as it is in our function, the x-axis (y=0) is a horizontal asymptote. So, for our function, y = 0 is the horizontal asymptote.
Key Concepts
Domain of a FunctionVertical AsymptotesHorizontal Asymptotes
Domain of a Function
Understanding the domain of a rational function is crucial, as it tells us the set of all possible 'x' values for which the function is defined. In the given function, \( f(x) = \frac{2x + 7}{2x^{2} + 5x - 3} \), the domain consists of all real numbers except those that make the denominator zero, because division by zero is undefined.
To find these exclusions, we set the denominator, \(2x^{2} + 5x - 3\), equal to zero and solve for x. Using the quadratic formula, we find that \(x = 0.5\) and \(x = -3\) are the roots. Thus, these values are excluded from the domain.
Therefore, the domain of the function can be expressed as all real numbers except \(x = 0.5\) and \(x = -3\). This can also be stated in interval notation as:\
To find these exclusions, we set the denominator, \(2x^{2} + 5x - 3\), equal to zero and solve for x. Using the quadratic formula, we find that \(x = 0.5\) and \(x = -3\) are the roots. Thus, these values are excluded from the domain.
Therefore, the domain of the function can be expressed as all real numbers except \(x = 0.5\) and \(x = -3\). This can also be stated in interval notation as:\
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- \((-\infty, -3) \cup (-3, 0.5) \cup (0.5, \infty)\) \
Vertical Asymptotes
Vertical asymptotes in a rational function occur at certain 'x' values where the function heads towards positive or negative infinity, effectively making the function undefined at those points.
For the function \( f(x) = \frac{2x + 7}{2x^{2} + 5x - 3} \), we already identified that \(x = 0.5\) and \(x = -3\), obtained from the zeros of the denominator in the domain analysis, are potential values for vertical asymptotes.
These vertical lines, \(x = 0.5\) and \(x = -3\), are where the function sharply increases or decreases without crossing. Therefore, at these points, the function has vertical asymptotes.
For the function \( f(x) = \frac{2x + 7}{2x^{2} + 5x - 3} \), we already identified that \(x = 0.5\) and \(x = -3\), obtained from the zeros of the denominator in the domain analysis, are potential values for vertical asymptotes.
These vertical lines, \(x = 0.5\) and \(x = -3\), are where the function sharply increases or decreases without crossing. Therefore, at these points, the function has vertical asymptotes.
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- These asymptotes divide the graph into separate regions. \
- The presence of these asymptotes often indicates that the graph will have different behaviors on each side of the asymptote. \
Horizontal Asymptotes
Horizontal asymptotes describe the behavior of a rational function as the input variable 'x' approaches infinity or negative infinity. They provide insight into the function's end behavior.
For the given function \( f(x) = \frac{2x + 7}{2x^{2} + 5x - 3} \), compare the degrees of the polynomial in the numerator and denominator. Here, the numerator is degree 1 and the denominator is degree 2.
Understanding horizontal asymptotes helps predict long-term trends in the behavior of the function, a concept particularly useful in real-world applications like analyzing growth models or rates of change.
For the given function \( f(x) = \frac{2x + 7}{2x^{2} + 5x - 3} \), compare the degrees of the polynomial in the numerator and denominator. Here, the numerator is degree 1 and the denominator is degree 2.
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- When the degree of the numerator is less than the degree of the denominator, the x-axis (y = 0) is the horizontal asymptote. \
- This shows that as 'x' moves towards infinity or negative infinity, the function values approach zero. \
Understanding horizontal asymptotes helps predict long-term trends in the behavior of the function, a concept particularly useful in real-world applications like analyzing growth models or rates of change.
Other exercises in this chapter
Problem 17
Show that the given value of \(x\) is a zero of the polynomial. Use the zero to completely factor the polynomial. $$p(x)=3 x^{3}+x^{2}+24 x+8 ; x=-\frac{1}{3}$$
View solution Problem 17
Find all the zeros, real and nonreal, of the polynomial. Then express \(p(x)\) as a product of linear factors. $$p(x)=x^{2}-\sqrt{3}$$
View solution Problem 17
Write each polynomial in the form \(p(x)=d(x) q(x)+r(x),\) where \(p(x)\) is the given polynomial and \(d(x)\) is the given factor. You may use synthetic divisi
View solution Problem 17
Determine what type of symmetry, if any, the function illustrates. Classify the function as odd, even, or neither. $$f(x)=-3 x^{3}+1$$
View solution