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Sagot :
To graph the function [tex]\( c(x) = \frac{7}{x^2 - 9} \)[/tex], we need to consider several key points, including the vertical asymptotes, horizontal asymptotes, and the behavior of the function around these asymptotes. Let's go step-by-step.
### Step 1: Identify the Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is zero and the numerator is non-zero. Set the denominator equal to zero:
[tex]\[ x^2 - 9 = 0 \][/tex]
[tex]\[ x^2 = 9 \][/tex]
[tex]\[ x = \pm 3 \][/tex]
So, the vertical asymptotes are at [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex].
### Step 2: Identify the Horizontal Asymptote
To find the horizontal asymptote, we analyze the behavior of the function as [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex]. As [tex]\( x \to \pm \infty \)[/tex], the term [tex]\( x^2 \)[/tex] grows much larger than 9, so:
[tex]\[ c(x) = \frac{7}{x^2 - 9} \approx \frac{7}{x^2} \][/tex]
Thus, as [tex]\( x \to \pm \infty \)[/tex], [tex]\( c(x) \to 0 \)[/tex]. The horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
### Step 3: Plot Points Around Asymptotes
Let's calculate values of [tex]\( c(x) \)[/tex] for points around the vertical asymptotes:
#### Points to the left of [tex]\( x = -3 \)[/tex]:
- [tex]\( x = -4 \)[/tex]:
[tex]\[ c(-4) = \frac{7}{(-4)^2 - 9} = \frac{7}{16 - 9} = \frac{7}{7} = 1 \][/tex]
- [tex]\( x = -5 \)[/tex]:
[tex]\[ c(-5) = \frac{7}{(-5)^2 - 9} = \frac{7}{25 - 9} = \frac{7}{16} \approx 0.4375 \][/tex]
#### Points to the right of [tex]\( x = -3 \)[/tex] but to the left of [tex]\( x = 3 \)[/tex]:
- [tex]\( x = -2 \)[/tex]:
[tex]\[ c(-2) = \frac{7}{(-2)^2 - 9} = \frac{7}{4 - 9} = \frac{7}{-5} = -1.4 \][/tex]
- [tex]\( x = 0 \)[/tex]:
[tex]\[ c(0) = \frac{7}{0^2 - 9} = \frac{7}{-9} = -0.777\][/tex]
#### Points to the right of [tex]\( x = 3 \)[/tex]:
- [tex]\( x = 4 \)[/tex]:
[tex]\[ c(4) = \frac{7}{4^2 - 9} = \frac{7}{16 - 9} = \frac{7}{7} = 1 \][/tex]
- [tex]\( x = 5 \)[/tex]:
[tex]\[ c(5) = \frac{7}{5^2 - 9} = \frac{7}{25 - 9} = \frac{7}{16} \approx 0.4375 \][/tex]
### Step 4: Sketch the Graph
Now that we have key points and asymptotes, we are ready to sketch the graph.
1. Draw vertical lines at [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex] to represent the vertical asymptotes.
2. Draw a horizontal line along [tex]\( y = 0 \)[/tex] to represent the horizontal asymptote.
3. Plot the calculated points:
- [tex]\((-5, 0.4375)\)[/tex], [tex]\((-4, 1)\)[/tex] to the left of [tex]\( x = -3 \)[/tex].
- [tex]\((-2, -1.4)\)[/tex], [tex]\((0, -0.777)\)[/tex] between [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex].
- [tex]\((4, 1)\)[/tex], [tex]\((5, 0.4375)\)[/tex] to the right of [tex]\( x = 3 \)[/tex].
4. Connect the points with smooth curves, noting that the function approaches the asymptotes as it gets closer to [tex]\( x = \pm 3 \)[/tex] and [tex]\( y = 0 \)[/tex].
Your graph should show that [tex]\( c(x) \)[/tex] grows large positively or negatively near the vertical asymptotes and approaches zero as [tex]\( x \rightarrow \pm \infty \)[/tex].
### Step 1: Identify the Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is zero and the numerator is non-zero. Set the denominator equal to zero:
[tex]\[ x^2 - 9 = 0 \][/tex]
[tex]\[ x^2 = 9 \][/tex]
[tex]\[ x = \pm 3 \][/tex]
So, the vertical asymptotes are at [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex].
### Step 2: Identify the Horizontal Asymptote
To find the horizontal asymptote, we analyze the behavior of the function as [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex]. As [tex]\( x \to \pm \infty \)[/tex], the term [tex]\( x^2 \)[/tex] grows much larger than 9, so:
[tex]\[ c(x) = \frac{7}{x^2 - 9} \approx \frac{7}{x^2} \][/tex]
Thus, as [tex]\( x \to \pm \infty \)[/tex], [tex]\( c(x) \to 0 \)[/tex]. The horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
### Step 3: Plot Points Around Asymptotes
Let's calculate values of [tex]\( c(x) \)[/tex] for points around the vertical asymptotes:
#### Points to the left of [tex]\( x = -3 \)[/tex]:
- [tex]\( x = -4 \)[/tex]:
[tex]\[ c(-4) = \frac{7}{(-4)^2 - 9} = \frac{7}{16 - 9} = \frac{7}{7} = 1 \][/tex]
- [tex]\( x = -5 \)[/tex]:
[tex]\[ c(-5) = \frac{7}{(-5)^2 - 9} = \frac{7}{25 - 9} = \frac{7}{16} \approx 0.4375 \][/tex]
#### Points to the right of [tex]\( x = -3 \)[/tex] but to the left of [tex]\( x = 3 \)[/tex]:
- [tex]\( x = -2 \)[/tex]:
[tex]\[ c(-2) = \frac{7}{(-2)^2 - 9} = \frac{7}{4 - 9} = \frac{7}{-5} = -1.4 \][/tex]
- [tex]\( x = 0 \)[/tex]:
[tex]\[ c(0) = \frac{7}{0^2 - 9} = \frac{7}{-9} = -0.777\][/tex]
#### Points to the right of [tex]\( x = 3 \)[/tex]:
- [tex]\( x = 4 \)[/tex]:
[tex]\[ c(4) = \frac{7}{4^2 - 9} = \frac{7}{16 - 9} = \frac{7}{7} = 1 \][/tex]
- [tex]\( x = 5 \)[/tex]:
[tex]\[ c(5) = \frac{7}{5^2 - 9} = \frac{7}{25 - 9} = \frac{7}{16} \approx 0.4375 \][/tex]
### Step 4: Sketch the Graph
Now that we have key points and asymptotes, we are ready to sketch the graph.
1. Draw vertical lines at [tex]\( x = 3 \)[/tex] and [tex]\( x = -3 \)[/tex] to represent the vertical asymptotes.
2. Draw a horizontal line along [tex]\( y = 0 \)[/tex] to represent the horizontal asymptote.
3. Plot the calculated points:
- [tex]\((-5, 0.4375)\)[/tex], [tex]\((-4, 1)\)[/tex] to the left of [tex]\( x = -3 \)[/tex].
- [tex]\((-2, -1.4)\)[/tex], [tex]\((0, -0.777)\)[/tex] between [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex].
- [tex]\((4, 1)\)[/tex], [tex]\((5, 0.4375)\)[/tex] to the right of [tex]\( x = 3 \)[/tex].
4. Connect the points with smooth curves, noting that the function approaches the asymptotes as it gets closer to [tex]\( x = \pm 3 \)[/tex] and [tex]\( y = 0 \)[/tex].
Your graph should show that [tex]\( c(x) \)[/tex] grows large positively or negatively near the vertical asymptotes and approaches zero as [tex]\( x \rightarrow \pm \infty \)[/tex].
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