IDNLearn.com is designed to help you find reliable answers to any question you have. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
Sagot :
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.
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
