Answer :
Sure! Let's walk through the process step by step for the given rational function:
[tex]\[ f(x) = \frac{5}{x-4} - 4 \][/tex]
### Step 1: Determine the Vertical Asymptote
Vertical asymptotes occur where the denominator of the rational function is zero, which causes the function to be undefined.
For [tex]\(\frac{5}{x-4}\)[/tex], the denominator is [tex]\(x-4\)[/tex].
Set the denominator equal to zero:
[tex]\[ x - 4 = 0 \][/tex]
[tex]\[ x = 4 \][/tex]
Thus, there is a vertical asymptote at [tex]\( x = 4 \)[/tex].
### Step 2: Determine the Horizontal Asymptote
Horizontal asymptotes are found by analyzing the behavior of the function as [tex]\( x \)[/tex] approaches infinity.
Consider [tex]\( \frac{5}{x-4} \)[/tex]. As [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex], [tex]\(\frac{5}{x-4} \)[/tex] approaches 0.
The horizontal asymptote, therefore, is determined by the constant term left after analyzing what happens to [tex]\(\frac{5}{x-4}\)[/tex] at infinity:
[tex]\[ \frac{5}{x-4} \to 0 \quad \text{as} \quad x \to \pm\infty \][/tex]
Thus:
[tex]\[ f(x) \to -4 \quad \text{as} \quad x \to \pm\infty \][/tex]
So, there is a horizontal asymptote at [tex]\( y = -4 \)[/tex].
### Step 3: Determine the [tex]\(x\)[/tex]-Intercept
The [tex]\(x\)[/tex]-intercept occurs where the function equals zero. Set [tex]\( f(x) \)[/tex] to 0 and solve for [tex]\( x \)[/tex].
[tex]\[ \frac{5}{x-4} - 4 = 0 \][/tex]
Solving this:
[tex]\[ \frac{5}{x-4} = 4 \][/tex]
[tex]\[ 5 = 4(x-4) \][/tex]
[tex]\[ 5 = 4x - 16 \][/tex]
[tex]\[ 4x = 21 \][/tex]
[tex]\[ x = \frac{21}{4} = 5.25 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercept is at [tex]\( x = \frac{21}{4} \)[/tex].
### Step 4: Determine the [tex]\(y\)[/tex]-Intercept
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{5}{0-4} - 4 \][/tex]
[tex]\[ f(0) = \frac{5}{-4} - 4 \][/tex]
[tex]\[ f(0) = -\frac{5}{4} - 4 \][/tex]
[tex]\[ f(0) = -\frac{5}{4} - \frac{16}{4} \][/tex]
[tex]\[ f(0) = -\frac{21}{4} \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is at [tex]\( y = -\frac{21}{4} \)[/tex].
### Plotting the Function
We can now plot the intercepts and asymptotes on the graph of the function [tex]\( f(x) = \frac{5}{x-4} - 4 \)[/tex].
1. Vertical Asymptote: Draw a dashed vertical line at [tex]\( x = 4 \)[/tex].
2. Horizontal Asymptote: Draw a dashed horizontal line at [tex]\( y = -4 \)[/tex].
3. [tex]\( x \)[/tex]-Intercept: Plot the point [tex]\( \left( \frac{21}{4}, 0 \right) \)[/tex] which is approximately [tex]\( (5.25, 0) \)[/tex].
4. [tex]\( y \)[/tex]-Intercept: Plot the point [tex]\( \left( 0, -\frac{21}{4} \right) \)[/tex] which is approximately [tex]\( (0, -5.25) \)[/tex].
Here’s a basic sketch representing the function along with its intercepts and asymptotes:
[tex]\[ \begin{array}{c} \begin{tikzpicture}[scale=1] % Axes \draw[black!50] (-2,-6) grid (8,2); \draw[thick, ->] (-2,0) -- (8,0) node[right] {\(x\)}; \draw[thick, ->] (0,-6) -- (0,2) node[above] {\(y\)}; % Asymptotes \draw[red, dashed] (4,-6) -- (4,2); \node at (4,-6.5) [red] {\(x=4\)}; \draw[green, dashed] (-2,-4) -- (8,-4); \node at (-2.5,-4.0) [green] {\(y=-4\)}; % Intercepts \draw[blue, fill=blue] (5.25,0) circle [radius=0.05]; \node at (5.35,0) [below] {\( (\frac{21}{4}, 0) \)}; \draw[magenta, fill=magenta] (0,-5.25) circle [radius=0.05]; \node at (0,-5.6) [left] { \( (0, -\frac{21}{4}) \) }; % Function plot \draw[cyan, thick] plot [domain=-2:3.8, samples=100] (\x,{5/(\x-4)-4}); \draw[cyan, thick] plot [domain=4.2:8, samples=100] (\x,{5/(\x-4)-4}); \node at (5,-3) [cyan] {\(f(x) = \frac{5}{x-4} - 4\)}; \end{tikzpicture} \end{array} \][/tex]
And there you have it! The graph of the function, its asymptotes, and intercepts plotted.
[tex]\[ f(x) = \frac{5}{x-4} - 4 \][/tex]
### Step 1: Determine the Vertical Asymptote
Vertical asymptotes occur where the denominator of the rational function is zero, which causes the function to be undefined.
For [tex]\(\frac{5}{x-4}\)[/tex], the denominator is [tex]\(x-4\)[/tex].
Set the denominator equal to zero:
[tex]\[ x - 4 = 0 \][/tex]
[tex]\[ x = 4 \][/tex]
Thus, there is a vertical asymptote at [tex]\( x = 4 \)[/tex].
### Step 2: Determine the Horizontal Asymptote
Horizontal asymptotes are found by analyzing the behavior of the function as [tex]\( x \)[/tex] approaches infinity.
Consider [tex]\( \frac{5}{x-4} \)[/tex]. As [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex], [tex]\(\frac{5}{x-4} \)[/tex] approaches 0.
The horizontal asymptote, therefore, is determined by the constant term left after analyzing what happens to [tex]\(\frac{5}{x-4}\)[/tex] at infinity:
[tex]\[ \frac{5}{x-4} \to 0 \quad \text{as} \quad x \to \pm\infty \][/tex]
Thus:
[tex]\[ f(x) \to -4 \quad \text{as} \quad x \to \pm\infty \][/tex]
So, there is a horizontal asymptote at [tex]\( y = -4 \)[/tex].
### Step 3: Determine the [tex]\(x\)[/tex]-Intercept
The [tex]\(x\)[/tex]-intercept occurs where the function equals zero. Set [tex]\( f(x) \)[/tex] to 0 and solve for [tex]\( x \)[/tex].
[tex]\[ \frac{5}{x-4} - 4 = 0 \][/tex]
Solving this:
[tex]\[ \frac{5}{x-4} = 4 \][/tex]
[tex]\[ 5 = 4(x-4) \][/tex]
[tex]\[ 5 = 4x - 16 \][/tex]
[tex]\[ 4x = 21 \][/tex]
[tex]\[ x = \frac{21}{4} = 5.25 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercept is at [tex]\( x = \frac{21}{4} \)[/tex].
### Step 4: Determine the [tex]\(y\)[/tex]-Intercept
The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{5}{0-4} - 4 \][/tex]
[tex]\[ f(0) = \frac{5}{-4} - 4 \][/tex]
[tex]\[ f(0) = -\frac{5}{4} - 4 \][/tex]
[tex]\[ f(0) = -\frac{5}{4} - \frac{16}{4} \][/tex]
[tex]\[ f(0) = -\frac{21}{4} \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is at [tex]\( y = -\frac{21}{4} \)[/tex].
### Plotting the Function
We can now plot the intercepts and asymptotes on the graph of the function [tex]\( f(x) = \frac{5}{x-4} - 4 \)[/tex].
1. Vertical Asymptote: Draw a dashed vertical line at [tex]\( x = 4 \)[/tex].
2. Horizontal Asymptote: Draw a dashed horizontal line at [tex]\( y = -4 \)[/tex].
3. [tex]\( x \)[/tex]-Intercept: Plot the point [tex]\( \left( \frac{21}{4}, 0 \right) \)[/tex] which is approximately [tex]\( (5.25, 0) \)[/tex].
4. [tex]\( y \)[/tex]-Intercept: Plot the point [tex]\( \left( 0, -\frac{21}{4} \right) \)[/tex] which is approximately [tex]\( (0, -5.25) \)[/tex].
Here’s a basic sketch representing the function along with its intercepts and asymptotes:
[tex]\[ \begin{array}{c} \begin{tikzpicture}[scale=1] % Axes \draw[black!50] (-2,-6) grid (8,2); \draw[thick, ->] (-2,0) -- (8,0) node[right] {\(x\)}; \draw[thick, ->] (0,-6) -- (0,2) node[above] {\(y\)}; % Asymptotes \draw[red, dashed] (4,-6) -- (4,2); \node at (4,-6.5) [red] {\(x=4\)}; \draw[green, dashed] (-2,-4) -- (8,-4); \node at (-2.5,-4.0) [green] {\(y=-4\)}; % Intercepts \draw[blue, fill=blue] (5.25,0) circle [radius=0.05]; \node at (5.35,0) [below] {\( (\frac{21}{4}, 0) \)}; \draw[magenta, fill=magenta] (0,-5.25) circle [radius=0.05]; \node at (0,-5.6) [left] { \( (0, -\frac{21}{4}) \) }; % Function plot \draw[cyan, thick] plot [domain=-2:3.8, samples=100] (\x,{5/(\x-4)-4}); \draw[cyan, thick] plot [domain=4.2:8, samples=100] (\x,{5/(\x-4)-4}); \node at (5,-3) [cyan] {\(f(x) = \frac{5}{x-4} - 4\)}; \end{tikzpicture} \end{array} \][/tex]
And there you have it! The graph of the function, its asymptotes, and intercepts plotted.
