Answer :
To graph the rational function
[tex]\[ f(x) = \frac{-2x - 6}{x^2 + 4x + 3} \][/tex]
we need to perform the following steps: identify the asymptotes and holes, then plot several points to understand the shape of the graph.
### Step 1: Identify Vertical Asymptotes and Holes
1. Vertical Asymptotes: These occur where the denominator is equal to zero (and not canceled out by the numerator, which would indicate a hole).
The denominator is [tex]\( x^2 + 4x + 3 \)[/tex]. We set it to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 + 4x + 3 = 0 \][/tex]
Factoring the quadratic expression, we get:
[tex]\[ (x + 1)(x + 3) = 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] and [tex]\( x = -3 \)[/tex].
Vertical asymptotes are at [tex]\( x = -1 \)[/tex] and [tex]\( x = -3 \)[/tex], provided these are not canceled by zeros in the numerator.
2. Holes: Holes in the graph occur where both the numerator and denominator are zero.
We analyze the numerator [tex]\( -2x - 6 \)[/tex] and set it to zero:
[tex]\[ -2x - 6 = 0 \implies x = -3 \][/tex]
Since [tex]\( x = -3 \)[/tex] is also a root of the denominator, we must check for a hole at [tex]\( x = -3 \)[/tex].
Therefore, we simplify the original function by factoring and canceling common factors:
[tex]\[ f(x) = \frac{-2(x + 3)}{(x + 1)(x + 3)} \][/tex]
After canceling out the common factor [tex]\( x + 3 \)[/tex]:
[tex]\[ f(x) = \frac{-2}{x + 1}, \quad \text{for } x \neq -3 \][/tex]
This confirms that there is a hole at [tex]\( x = -3 \)[/tex].
### Step 2: Identify the Horizontal Asymptote
Horizontal asymptotes are determined by comparing the degrees of the numerator and the denominator.
- The degree of the numerator [tex]\(-2x - 6\)[/tex] is 1.
- The degree of the denominator [tex]\(x^2 + 4x + 3\)[/tex] is 2.
Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
### Step 3: Plot Two Points for Each Piece
1. For [tex]\( x < -3 \)[/tex]:
[tex]\[ f(x) = \frac{-2}{x + 1} \][/tex]
Choose [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = \frac{-2}{-4 + 1} = \frac{-2}{-3} = \frac{2}{3} \][/tex]
Choose [tex]\( x = -5 \)[/tex]:
[tex]\[ f(-5) = \frac{-2}{-5 + 1} = \frac{-2}{-4} = \frac{1}{2} \][/tex]
2. For [tex]\( -3 < x < -1 \)[/tex]:
Choose [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \frac{-2}{-2 + 1} = \frac{-2}{-1} = 2 \][/tex]
Choose [tex]\( x = -2.5 \)[/tex]:
[tex]\[ f(-2.5) = \frac{-2}{-2.5 + 1} = \frac{-2}{-1.5} = \frac{4}{3} \][/tex]
3. For [tex]\( x > -1 \)[/tex]:
Choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{-2}{0 + 1} = -2 \][/tex]
Choose [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{-2}{1 + 1} = \frac{-2}{2} = -1 \][/tex]
### Step 4: Draw the Graph
1. Vertical asymptotes at [tex]\( x = -1 \)[/tex] and the hole at [tex]\( x = -3 \)[/tex] (draw dashed vertical lines at these points).
2. Horizontal asymptote at [tex]\( y = 0 \)[/tex] (draw a dashed horizontal line at [tex]\( y = 0 \)[/tex]).
3. Plot the points identified and draw the curves, ensuring the graph approaches the vertical and horizontal asymptotes but doesn't cross them.
Finally, plot these points and draw the function. Use a hollow dot at the hole [tex]\( x = -3 \)[/tex] to indicate the discontinuity there.
Here's an outline of what the graph should look like by hand-drawing the asymptotes, points, and ensuring the appropriate behavior near the asymptotes.
[tex]\[ f(x) = \frac{-2x - 6}{x^2 + 4x + 3} \][/tex]
we need to perform the following steps: identify the asymptotes and holes, then plot several points to understand the shape of the graph.
### Step 1: Identify Vertical Asymptotes and Holes
1. Vertical Asymptotes: These occur where the denominator is equal to zero (and not canceled out by the numerator, which would indicate a hole).
The denominator is [tex]\( x^2 + 4x + 3 \)[/tex]. We set it to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 + 4x + 3 = 0 \][/tex]
Factoring the quadratic expression, we get:
[tex]\[ (x + 1)(x + 3) = 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] and [tex]\( x = -3 \)[/tex].
Vertical asymptotes are at [tex]\( x = -1 \)[/tex] and [tex]\( x = -3 \)[/tex], provided these are not canceled by zeros in the numerator.
2. Holes: Holes in the graph occur where both the numerator and denominator are zero.
We analyze the numerator [tex]\( -2x - 6 \)[/tex] and set it to zero:
[tex]\[ -2x - 6 = 0 \implies x = -3 \][/tex]
Since [tex]\( x = -3 \)[/tex] is also a root of the denominator, we must check for a hole at [tex]\( x = -3 \)[/tex].
Therefore, we simplify the original function by factoring and canceling common factors:
[tex]\[ f(x) = \frac{-2(x + 3)}{(x + 1)(x + 3)} \][/tex]
After canceling out the common factor [tex]\( x + 3 \)[/tex]:
[tex]\[ f(x) = \frac{-2}{x + 1}, \quad \text{for } x \neq -3 \][/tex]
This confirms that there is a hole at [tex]\( x = -3 \)[/tex].
### Step 2: Identify the Horizontal Asymptote
Horizontal asymptotes are determined by comparing the degrees of the numerator and the denominator.
- The degree of the numerator [tex]\(-2x - 6\)[/tex] is 1.
- The degree of the denominator [tex]\(x^2 + 4x + 3\)[/tex] is 2.
Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
### Step 3: Plot Two Points for Each Piece
1. For [tex]\( x < -3 \)[/tex]:
[tex]\[ f(x) = \frac{-2}{x + 1} \][/tex]
Choose [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = \frac{-2}{-4 + 1} = \frac{-2}{-3} = \frac{2}{3} \][/tex]
Choose [tex]\( x = -5 \)[/tex]:
[tex]\[ f(-5) = \frac{-2}{-5 + 1} = \frac{-2}{-4} = \frac{1}{2} \][/tex]
2. For [tex]\( -3 < x < -1 \)[/tex]:
Choose [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \frac{-2}{-2 + 1} = \frac{-2}{-1} = 2 \][/tex]
Choose [tex]\( x = -2.5 \)[/tex]:
[tex]\[ f(-2.5) = \frac{-2}{-2.5 + 1} = \frac{-2}{-1.5} = \frac{4}{3} \][/tex]
3. For [tex]\( x > -1 \)[/tex]:
Choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{-2}{0 + 1} = -2 \][/tex]
Choose [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{-2}{1 + 1} = \frac{-2}{2} = -1 \][/tex]
### Step 4: Draw the Graph
1. Vertical asymptotes at [tex]\( x = -1 \)[/tex] and the hole at [tex]\( x = -3 \)[/tex] (draw dashed vertical lines at these points).
2. Horizontal asymptote at [tex]\( y = 0 \)[/tex] (draw a dashed horizontal line at [tex]\( y = 0 \)[/tex]).
3. Plot the points identified and draw the curves, ensuring the graph approaches the vertical and horizontal asymptotes but doesn't cross them.
Finally, plot these points and draw the function. Use a hollow dot at the hole [tex]\( x = -3 \)[/tex] to indicate the discontinuity there.
Here's an outline of what the graph should look like by hand-drawing the asymptotes, points, and ensuring the appropriate behavior near the asymptotes.
