Answer :
To graph the function [tex]\( q(x) = \frac{-4}{2x - 3} \)[/tex] and identify its asymptotes, we need to follow a systematic approach. Let's break it down step-by-step:
### Step 1: Identify the Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is equal to zero because the function value tends towards infinity at these points.
For [tex]\( q(x) = \frac{-4}{2x - 3} \)[/tex]:
Set the denominator equal to zero:
[tex]\[ 2x - 3 = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 3 \][/tex]
[tex]\[ x = \frac{3}{2} \][/tex]
So, there is a vertical asymptote at [tex]\( x = \frac{3}{2} \)[/tex].
### Step 2: Identify the Horizontal Asymptote
Horizontal asymptotes are identified by analyzing the behavior of the function as [tex]\( x \)[/tex] approaches positive or negative infinity.
For the function [tex]\( q(x) = \frac{-4}{2x - 3} \)[/tex]:
Consider the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{-4}{2x - 3} \][/tex]
As [tex]\( x \)[/tex] becomes very large, the term [tex]\( -3 \)[/tex] in the denominator becomes negligible. Thus, the function approaches:
[tex]\[ \lim_{x \to \infty} \frac{-4}{2x - 3} \approx \frac{-4}{2x} \][/tex]
[tex]\[ = \lim_{x \to \infty} \frac{-4}{2} \cdot \frac{1}{x} \][/tex]
[tex]\[ = \frac{-4}{2} \cdot 0 \][/tex]
[tex]\[ = 0 \][/tex]
So, the horizontal asymptote is at [tex]\( y = 0 \)[/tex].
### Step 3: Graph the Function
To plot the function, we follow these steps:
1. Plot the Asymptotes:
- Draw a vertical line at [tex]\( x = \frac{3}{2} \)[/tex].
- Draw a horizontal line at [tex]\( y = 0 \)[/tex].
2. Plot the Function:
- Choose several points on either side of the vertical asymptote to get the shape of the curve.
- As [tex]\( x \)[/tex] approaches the vertical asymptote from either side, the function value tends towards [tex]\( \pm \infty \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( \pm \infty \)[/tex], the function approaches the horizontal asymptote [tex]\( y = 0 \)[/tex].
Here is a sketch of the graph based on these observations:
### Graph of [tex]\( q(x) = \frac{-4}{2x - 3} \)[/tex]:
1. Vertical Asymptote: [tex]\( x = \frac{3}{2} \)[/tex]
2. Horizontal Asymptote: [tex]\( y = 0 \)[/tex]
[tex]\[ \begin{array}{c|c} x & q(x) \\ \hline -2 & \frac{-4}{-4 - 3} = \frac{-4}{-7} \approx 0.571 \\ 0 & \frac{-4}{-3} \approx 1.333 \\ 1 & \frac{-4}{-1} = 4 \\ \end{array} \][/tex]
As [tex]\( x \to \frac{3}{2}^- \)[/tex], [tex]\( q(x) \to -\infty \)[/tex].
As [tex]\( x \to \frac{3}{2}^+ \)[/tex], [tex]\( q(x) \to \infty \)[/tex].
### Drawing:
On a Cartesian plane, draw the vertical line [tex]\( x = \frac{3}{2} \)[/tex] and the horizontal line [tex]\( y = 0 \)[/tex]. Plot points such as [tex]\( (-2, 0.571) \)[/tex], [tex]\( (0, 1.333) \)[/tex], and [tex]\( (1, 4) \)[/tex]. Draw the curve approaching the asymptotes:
- For [tex]\( x < \frac{3}{2} \)[/tex], the curve falls from positive values approaching the vertical asymptote.
- For [tex]\( x > \frac{3}{2} \)[/tex], the curve rises from negative values approaching the vertical asymptote.
In this manner, the graph will display the behavior defined by the asymptotes and calculated points.
### Step 1: Identify the Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is equal to zero because the function value tends towards infinity at these points.
For [tex]\( q(x) = \frac{-4}{2x - 3} \)[/tex]:
Set the denominator equal to zero:
[tex]\[ 2x - 3 = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 3 \][/tex]
[tex]\[ x = \frac{3}{2} \][/tex]
So, there is a vertical asymptote at [tex]\( x = \frac{3}{2} \)[/tex].
### Step 2: Identify the Horizontal Asymptote
Horizontal asymptotes are identified by analyzing the behavior of the function as [tex]\( x \)[/tex] approaches positive or negative infinity.
For the function [tex]\( q(x) = \frac{-4}{2x - 3} \)[/tex]:
Consider the limit as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{-4}{2x - 3} \][/tex]
As [tex]\( x \)[/tex] becomes very large, the term [tex]\( -3 \)[/tex] in the denominator becomes negligible. Thus, the function approaches:
[tex]\[ \lim_{x \to \infty} \frac{-4}{2x - 3} \approx \frac{-4}{2x} \][/tex]
[tex]\[ = \lim_{x \to \infty} \frac{-4}{2} \cdot \frac{1}{x} \][/tex]
[tex]\[ = \frac{-4}{2} \cdot 0 \][/tex]
[tex]\[ = 0 \][/tex]
So, the horizontal asymptote is at [tex]\( y = 0 \)[/tex].
### Step 3: Graph the Function
To plot the function, we follow these steps:
1. Plot the Asymptotes:
- Draw a vertical line at [tex]\( x = \frac{3}{2} \)[/tex].
- Draw a horizontal line at [tex]\( y = 0 \)[/tex].
2. Plot the Function:
- Choose several points on either side of the vertical asymptote to get the shape of the curve.
- As [tex]\( x \)[/tex] approaches the vertical asymptote from either side, the function value tends towards [tex]\( \pm \infty \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( \pm \infty \)[/tex], the function approaches the horizontal asymptote [tex]\( y = 0 \)[/tex].
Here is a sketch of the graph based on these observations:
### Graph of [tex]\( q(x) = \frac{-4}{2x - 3} \)[/tex]:
1. Vertical Asymptote: [tex]\( x = \frac{3}{2} \)[/tex]
2. Horizontal Asymptote: [tex]\( y = 0 \)[/tex]
[tex]\[ \begin{array}{c|c} x & q(x) \\ \hline -2 & \frac{-4}{-4 - 3} = \frac{-4}{-7} \approx 0.571 \\ 0 & \frac{-4}{-3} \approx 1.333 \\ 1 & \frac{-4}{-1} = 4 \\ \end{array} \][/tex]
As [tex]\( x \to \frac{3}{2}^- \)[/tex], [tex]\( q(x) \to -\infty \)[/tex].
As [tex]\( x \to \frac{3}{2}^+ \)[/tex], [tex]\( q(x) \to \infty \)[/tex].
### Drawing:
On a Cartesian plane, draw the vertical line [tex]\( x = \frac{3}{2} \)[/tex] and the horizontal line [tex]\( y = 0 \)[/tex]. Plot points such as [tex]\( (-2, 0.571) \)[/tex], [tex]\( (0, 1.333) \)[/tex], and [tex]\( (1, 4) \)[/tex]. Draw the curve approaching the asymptotes:
- For [tex]\( x < \frac{3}{2} \)[/tex], the curve falls from positive values approaching the vertical asymptote.
- For [tex]\( x > \frac{3}{2} \)[/tex], the curve rises from negative values approaching the vertical asymptote.
In this manner, the graph will display the behavior defined by the asymptotes and calculated points.
