Answer :
To graph the function [tex]\( f(x) = \frac{2}{x+1} - 3 \)[/tex], we'll follow several steps to understand its behavior and sketch its graph.
1. Identify Key Features:
- Vertical Asymptote: The function [tex]\( f(x) = \frac{2}{x+1} - 3 \)[/tex] has a vertical asymptote where the denominator is zero. The denominator [tex]\( x+1 \)[/tex] is zero when [tex]\( x = -1 \)[/tex]. So, there is a vertical asymptote at [tex]\( x = -1 \)[/tex].
- Horizontal Asymptote: As [tex]\( x \)[/tex] goes to [tex]\(\pm\infty\)[/tex], [tex]\( \frac{2}{x+1} \)[/tex] approaches zero, and thus [tex]\( f(x) \)[/tex] approaches [tex]\( -3 \)[/tex]. Therefore, there is a horizontal asymptote at [tex]\( y = -3 \)[/tex].
2. Intercepts:
- X-Intercept: Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{2}{x+1} - 3 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{2}{x+1} = 3 \][/tex]
[tex]\[ 2 = 3(x+1) \][/tex]
[tex]\[ 2 = 3x + 3 \][/tex]
[tex]\[ 3x = -1 \][/tex]
[tex]\[ x = -\frac{1}{3} \][/tex]
So, the x-intercept is at [tex]\( \left( -\frac{1}{3}, 0 \right) \)[/tex].
- Y-Intercept: Set [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{2}{0+1} - 3 = 2 - 3 = -1 \][/tex]
So, the y-intercept is at [tex]\( (0, -1) \)[/tex].
3. Behavior Near the Asymptotes:
- As [tex]\( x \)[/tex] approaches [tex]\( -1 \)[/tex] from the left ([tex]\( x \rightarrow -1^- \)[/tex]), [tex]\( \frac{2}{x+1} \)[/tex] becomes very large negatively, making [tex]\( f(x) \)[/tex] approach [tex]\( -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( -1 \)[/tex] from the right ([tex]\( x \rightarrow -1^+ \)[/tex]), [tex]\( \frac{2}{x+1} \)[/tex] becomes very large positively, making [tex]\( f(x) \)[/tex] approach [tex]\( \infty \)[/tex].
4. Plotting Points:
- To further understand the shape, we can evaluate the function at a few more points and plot them:
[tex]\[ f(-2) = \frac{2}{-2+1} - 3 = \frac{2}{-1} - 3 = -2 - 3 = -5 \][/tex]
[tex]\[ f(1) = \frac{2}{1+1} - 3 = \frac{2}{2} - 3 = 1 - 3 = -2 \][/tex]
5. Sketch the Graph:
- Draw vertical asymptote at [tex]\( x = -1 \)[/tex].
- Draw horizontal asymptote at [tex]\( y = -3 \)[/tex].
- Plot the intercepts: [tex]\( \left( -\frac{1}{3}, 0 \right) \)[/tex] and [tex]\( (0, -1) \)[/tex].
- Plot additional points like [tex]\( (-2, -5) \)[/tex] and [tex]\( (1, -2) \)[/tex].
By connecting these points and behavior patterns, the graph of [tex]\( f(x) = \frac{2}{x+1} - 3 \)[/tex] will look like this:
[tex]\[ \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = {$x$}, ylabel = {$f(x)$}, xmin=-3,xmax=3, ymin=-8,ymax=2, xtick={-3,-2,-1,0,1,2,3}, ytick={-8,-6,-4,-3,-2,0,1,2}, ] % Vertical asymptote \addplot[dashed, samples=2, domain=-3:3] {x==-1 ? -1e16 : 1/0}; % Horizontal asymptote \addplot[dashed, samples=2, domain=-3:3] {-3}; % The actual graph \addplot[domain=-3:-1.1, samples=100, thick] {2/(x+1) - 3}; \addplot[domain=-0.9:3, samples=100, thick] {2/(x+1) - 3}; % Adding key points \addplot[only marks, mark=*] coordinates{(-1/3, 0) (0, -1) (-2, -5) (1, -2)}; \end{axis} \end{tikzpicture} \][/tex]
1. Identify Key Features:
- Vertical Asymptote: The function [tex]\( f(x) = \frac{2}{x+1} - 3 \)[/tex] has a vertical asymptote where the denominator is zero. The denominator [tex]\( x+1 \)[/tex] is zero when [tex]\( x = -1 \)[/tex]. So, there is a vertical asymptote at [tex]\( x = -1 \)[/tex].
- Horizontal Asymptote: As [tex]\( x \)[/tex] goes to [tex]\(\pm\infty\)[/tex], [tex]\( \frac{2}{x+1} \)[/tex] approaches zero, and thus [tex]\( f(x) \)[/tex] approaches [tex]\( -3 \)[/tex]. Therefore, there is a horizontal asymptote at [tex]\( y = -3 \)[/tex].
2. Intercepts:
- X-Intercept: Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{2}{x+1} - 3 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{2}{x+1} = 3 \][/tex]
[tex]\[ 2 = 3(x+1) \][/tex]
[tex]\[ 2 = 3x + 3 \][/tex]
[tex]\[ 3x = -1 \][/tex]
[tex]\[ x = -\frac{1}{3} \][/tex]
So, the x-intercept is at [tex]\( \left( -\frac{1}{3}, 0 \right) \)[/tex].
- Y-Intercept: Set [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{2}{0+1} - 3 = 2 - 3 = -1 \][/tex]
So, the y-intercept is at [tex]\( (0, -1) \)[/tex].
3. Behavior Near the Asymptotes:
- As [tex]\( x \)[/tex] approaches [tex]\( -1 \)[/tex] from the left ([tex]\( x \rightarrow -1^- \)[/tex]), [tex]\( \frac{2}{x+1} \)[/tex] becomes very large negatively, making [tex]\( f(x) \)[/tex] approach [tex]\( -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( -1 \)[/tex] from the right ([tex]\( x \rightarrow -1^+ \)[/tex]), [tex]\( \frac{2}{x+1} \)[/tex] becomes very large positively, making [tex]\( f(x) \)[/tex] approach [tex]\( \infty \)[/tex].
4. Plotting Points:
- To further understand the shape, we can evaluate the function at a few more points and plot them:
[tex]\[ f(-2) = \frac{2}{-2+1} - 3 = \frac{2}{-1} - 3 = -2 - 3 = -5 \][/tex]
[tex]\[ f(1) = \frac{2}{1+1} - 3 = \frac{2}{2} - 3 = 1 - 3 = -2 \][/tex]
5. Sketch the Graph:
- Draw vertical asymptote at [tex]\( x = -1 \)[/tex].
- Draw horizontal asymptote at [tex]\( y = -3 \)[/tex].
- Plot the intercepts: [tex]\( \left( -\frac{1}{3}, 0 \right) \)[/tex] and [tex]\( (0, -1) \)[/tex].
- Plot additional points like [tex]\( (-2, -5) \)[/tex] and [tex]\( (1, -2) \)[/tex].
By connecting these points and behavior patterns, the graph of [tex]\( f(x) = \frac{2}{x+1} - 3 \)[/tex] will look like this:
[tex]\[ \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = {$x$}, ylabel = {$f(x)$}, xmin=-3,xmax=3, ymin=-8,ymax=2, xtick={-3,-2,-1,0,1,2,3}, ytick={-8,-6,-4,-3,-2,0,1,2}, ] % Vertical asymptote \addplot[dashed, samples=2, domain=-3:3] {x==-1 ? -1e16 : 1/0}; % Horizontal asymptote \addplot[dashed, samples=2, domain=-3:3] {-3}; % The actual graph \addplot[domain=-3:-1.1, samples=100, thick] {2/(x+1) - 3}; \addplot[domain=-0.9:3, samples=100, thick] {2/(x+1) - 3}; % Adding key points \addplot[only marks, mark=*] coordinates{(-1/3, 0) (0, -1) (-2, -5) (1, -2)}; \end{axis} \end{tikzpicture} \][/tex]
