Answer :
Let's follow the seven-step strategy to graph the rational function [tex]\( f(x) = \frac{4x^2}{x^2 - 4} \)[/tex].
Step 1: Determine the domain.
The domain of the function consists of all real numbers except where the denominator is zero. Setting the denominator [tex]\( x^2 - 4 = 0 \)[/tex] and solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 4 \][/tex]
[tex]\[ x = 2 \text{ or } x = -2 \][/tex]
Thus, the function is undefined at [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex]. Therefore, the domain is [tex]\( \mathbb{R} \setminus \{2, -2\} \)[/tex].
Step 2: Find the vertical asymptotes.
We identified the points where the denominator is zero in the previous step. Solving [tex]\( x^2 - 4 = 0 \)[/tex] gives us:
[tex]\[ x = 2 \][/tex]
[tex]\[ x = -2 \][/tex]
So, the vertical asymptotes are at:
[tex]\[ x = 2 \][/tex]
[tex]\[ x = -2 \][/tex]
Therefore, the correct answer is:
[tex]\[ \text{A. The equations of the vertical asymptotes are } x = 2, x = -2. \][/tex]
Step 3: Find the horizontal asymptote.
To find the horizontal asymptote, we need to analyze the degrees of the numerator and the denominator.
- The degree of the numerator [tex]\( 4x^2 \)[/tex] is 2.
- The degree of the denominator [tex]\( x^2 - 4 \)[/tex] is also 2.
Since the degrees of the numerator and the denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients. Here, the leading coefficient of the numerator is 4, and the leading coefficient of the denominator is 1. Therefore, the horizontal asymptote is:
[tex]\[ y = \frac{4}{1} = 4 \][/tex]
So, the correct answer is:
[tex]\[ \text{A. The equation of the horizontal asymptote is } y = 4. \][/tex]
Step 4: Plot points between and beyond each [tex]\( x \)[/tex]-intercept and vertical asymptote.
To find the values of the function at given values of [tex]\( x \)[/tex], we evaluate [tex]\( f(x) = \frac{4x^2}{x^2 - 4} \)[/tex] at the specified points:
[tex]\[ \begin{array}{c|c} x & f(x) \\ \hline -4 & f(-4) = \frac{4(-4)^2}{(-4)^2 - 4} = \frac{4 \cdot 16}{16 - 4} = \frac{64}{12} = \frac{16}{3} = 5.333333333333333 \\ -3 & f(-3) = \frac{4(-3)^2}{(-3)^2 - 4} = \frac{4 \cdot 9}{9 - 4} = \frac{36}{5} = 7.2 \\ -\frac{1}{2} & f\left(-\frac{1}{2}\right) = \frac{4\left(-\frac{1}{2}\right)^2}{\left(-\frac{1}{2}\right)^2 - 4} = \frac{4 \cdot \frac{1}{4}}{\frac{1}{4} - 4} = \frac{1}{-\frac{15}{4}} = -\frac{4}{15} = -0.26666666666666666 \\ \frac{1}{2} & f\left(\frac{1}{2}\right) = \frac{4\left(\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2 - 4} = \frac{4 \cdot \frac{1}{4}}{\frac{1}{4} - 4} = \frac{1}{-\frac{15}{4}} = -\frac{4}{15} = -0.26666666666666666 \\ 3 & f(3) = \frac{4(3)^2}{(3)^2 - 4} = \frac{4 \cdot 9}{9 - 4} = \frac{36}{5} = 7.2 \\ 4 & f(4) = \frac{4(4)^2}{(4)^2 - 4} = \frac{4 \cdot 16}{16 - 4} = \frac{64}{12} = \frac{16}{3} = 5.333333333333333 \\ \end{array} \][/tex]
So, we have the values:
[tex]\[ \begin{array}{ccccccc} x & -4 & -3 & -\frac{1}{2} & \frac{1}{2} & 3 & 4 \\ f(x)=\frac{4 x^2}{x^2-4} & \frac{16}{3} & \frac{36}{5} & -0.26666666666666666 & -0.26666666666666666 & 7.2 & \frac{16}{3} \\ \hline \end{array} \][/tex]
To summarize:
1. Vertical asymptotes: [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex]
2. Horizontal asymptote: [tex]\( y = 4 \)[/tex]
3. Function values at given points:
- [tex]\( f(-4) = \frac{16}{3} \)[/tex]
- [tex]\( f(-3) = \frac{36}{5} \)[/tex]
- [tex]\( f(-\frac{1}{2}) = -0.26666666666666666 \)[/tex]
- [tex]\( f(\frac{1}{2}) = -0.26666666666666666 \)[/tex]
- [tex]\( f(3) = 7.2 \)[/tex]
- [tex]\( f(4) = \frac{16}{3} \)[/tex]
This detailed information would help you graph the function [tex]\( f(x) = \frac{4x^2}{x^2 - 4} \)[/tex] correctly.
Step 1: Determine the domain.
The domain of the function consists of all real numbers except where the denominator is zero. Setting the denominator [tex]\( x^2 - 4 = 0 \)[/tex] and solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 4 \][/tex]
[tex]\[ x = 2 \text{ or } x = -2 \][/tex]
Thus, the function is undefined at [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex]. Therefore, the domain is [tex]\( \mathbb{R} \setminus \{2, -2\} \)[/tex].
Step 2: Find the vertical asymptotes.
We identified the points where the denominator is zero in the previous step. Solving [tex]\( x^2 - 4 = 0 \)[/tex] gives us:
[tex]\[ x = 2 \][/tex]
[tex]\[ x = -2 \][/tex]
So, the vertical asymptotes are at:
[tex]\[ x = 2 \][/tex]
[tex]\[ x = -2 \][/tex]
Therefore, the correct answer is:
[tex]\[ \text{A. The equations of the vertical asymptotes are } x = 2, x = -2. \][/tex]
Step 3: Find the horizontal asymptote.
To find the horizontal asymptote, we need to analyze the degrees of the numerator and the denominator.
- The degree of the numerator [tex]\( 4x^2 \)[/tex] is 2.
- The degree of the denominator [tex]\( x^2 - 4 \)[/tex] is also 2.
Since the degrees of the numerator and the denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients. Here, the leading coefficient of the numerator is 4, and the leading coefficient of the denominator is 1. Therefore, the horizontal asymptote is:
[tex]\[ y = \frac{4}{1} = 4 \][/tex]
So, the correct answer is:
[tex]\[ \text{A. The equation of the horizontal asymptote is } y = 4. \][/tex]
Step 4: Plot points between and beyond each [tex]\( x \)[/tex]-intercept and vertical asymptote.
To find the values of the function at given values of [tex]\( x \)[/tex], we evaluate [tex]\( f(x) = \frac{4x^2}{x^2 - 4} \)[/tex] at the specified points:
[tex]\[ \begin{array}{c|c} x & f(x) \\ \hline -4 & f(-4) = \frac{4(-4)^2}{(-4)^2 - 4} = \frac{4 \cdot 16}{16 - 4} = \frac{64}{12} = \frac{16}{3} = 5.333333333333333 \\ -3 & f(-3) = \frac{4(-3)^2}{(-3)^2 - 4} = \frac{4 \cdot 9}{9 - 4} = \frac{36}{5} = 7.2 \\ -\frac{1}{2} & f\left(-\frac{1}{2}\right) = \frac{4\left(-\frac{1}{2}\right)^2}{\left(-\frac{1}{2}\right)^2 - 4} = \frac{4 \cdot \frac{1}{4}}{\frac{1}{4} - 4} = \frac{1}{-\frac{15}{4}} = -\frac{4}{15} = -0.26666666666666666 \\ \frac{1}{2} & f\left(\frac{1}{2}\right) = \frac{4\left(\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2 - 4} = \frac{4 \cdot \frac{1}{4}}{\frac{1}{4} - 4} = \frac{1}{-\frac{15}{4}} = -\frac{4}{15} = -0.26666666666666666 \\ 3 & f(3) = \frac{4(3)^2}{(3)^2 - 4} = \frac{4 \cdot 9}{9 - 4} = \frac{36}{5} = 7.2 \\ 4 & f(4) = \frac{4(4)^2}{(4)^2 - 4} = \frac{4 \cdot 16}{16 - 4} = \frac{64}{12} = \frac{16}{3} = 5.333333333333333 \\ \end{array} \][/tex]
So, we have the values:
[tex]\[ \begin{array}{ccccccc} x & -4 & -3 & -\frac{1}{2} & \frac{1}{2} & 3 & 4 \\ f(x)=\frac{4 x^2}{x^2-4} & \frac{16}{3} & \frac{36}{5} & -0.26666666666666666 & -0.26666666666666666 & 7.2 & \frac{16}{3} \\ \hline \end{array} \][/tex]
To summarize:
1. Vertical asymptotes: [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex]
2. Horizontal asymptote: [tex]\( y = 4 \)[/tex]
3. Function values at given points:
- [tex]\( f(-4) = \frac{16}{3} \)[/tex]
- [tex]\( f(-3) = \frac{36}{5} \)[/tex]
- [tex]\( f(-\frac{1}{2}) = -0.26666666666666666 \)[/tex]
- [tex]\( f(\frac{1}{2}) = -0.26666666666666666 \)[/tex]
- [tex]\( f(3) = 7.2 \)[/tex]
- [tex]\( f(4) = \frac{16}{3} \)[/tex]
This detailed information would help you graph the function [tex]\( f(x) = \frac{4x^2}{x^2 - 4} \)[/tex] correctly.
