Math / Algebra

Questionon list
Follow the steps for graphing a rational function to graph the function $H(x)=\frac{4 x-24}{36-x^{2}}$ . tion y ion 10 on 11 n 12 13 14 15 16 7
B. The function has two vertical asymptotes. The leftmost asymptote is , and the rightmost asymptote is $\square$
(Type equations. Use integers or fractions for any numbers in the equations.) . The function has three vertical asymptotes. The leftmost asymptote is $\square$ , the middle asymptote is (Type equations. Use integers or fractions for any numbers in the equations.) D. There is no vertical asymptote. $\square$ and the rightmost asymptote is $\square$
Determine the hole, if it exists. Select the correct choice and, if necessary, fill in the answer box to complete your choice. There is a hole in the graph at the point $\left(6,-\frac{1}{3}\right)$ . $\square$ . (Type an ordered pair using integers or simplified fractions.) B. There are no holes in the graph.
Determine the behavior of the graph on either side of any vertical asymptotes, if any exist. Select the correct choice and, if necessary, fill in any answer box(es) to complete your choice. A. It approaches $\infty$ on one side of the asymptote(s) at $x=\square$ and $-\infty$ on the other. It approaches either $\infty$ or $-\infty$ on both sides of the asymptote(s) at $x=$ $\square$ . (Type integers or simplified fractions. Use a comma to separate answers as needed. Type each answer only once.) . It approaches $\infty$ on one side of the asymptote(s) at $x=-6$ and $-\infty$ on the other. (Type an integer or a simplified fraction. Use a comma to separate answers as needed. Type each answer only once.) . It approaches either $\infty$ or $-\infty$ on both sides of the asymptote(s) at $x=\square$ . $\square$ (Type an integer or a simplified fraction. Use a comma to separate answers as needed. Type each answer only once.) D. There is no vertical asymptote.
Determine the horizontal asymptote(s), if any exist. Select the correct choice and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one horizontal asymptote, $\square$ (Type an equation. Use integers or fractions for any numbers in the equation.) B. The function has two horizontal asymptotes. The top asymptote is $\square$ , and the bottom asymptote is $\square$ $\square$ . (Type equations. Use integers or fractions for any numbers in the equations.) C. There is no horizontal asymptote.

Studdy Solution

STEP 1

What is this asking? We need to graph $H(x) = \frac{4x - 24}{36 - x^2}$ and answer questions about its vertical asymptotes, holes, and horizontal asymptote. Watch out! Don't forget to simplify the function first!
Holes can be sneaky.

STEP 2

1. Simplify the Function
2. Find Vertical Asymptotes
3. Find Holes
4. Analyze Asymptotic Behavior
5. Find Horizontal Asymptote

STEP 3

We can factor out a $4$ from the numerator: $4x - 24 = 4(x - 6)$ .
So, $H(x) = \frac{4(x - 6)}{36 - x^2}$ .

STEP 4

The denominator is a difference of squares: $36 - x^2 = (6 - x)(6 + x)$ .
Now, $H(x) = \frac{4(x - 6)}{(6 - x)(6 + x)}$ .

STEP 5

Notice that $(x - 6)$ and $(6 - x)$ are almost the same!
We can rewrite $(6 - x)$ as $-1(x - 6)$ .
Then, $H(x) = \frac{4(x - 6)}{-1(x - 6)(6 + x)}$ .
Now we can divide the numerator and denominator by $(x - 6)$ , which simplifies to $1$ .
This gives us $H(x) = \frac{4}{-1(6 + x)} = \frac{-4}{x + 6}$ .

STEP 6

Vertical asymptotes happen when the denominator is zero and the numerator isn't also zero at the same $x$ value.
In our simplified function, the denominator is zero when $x + 6 = 0$ , which means $x = -6$ .

STEP 7

So, we have one vertical asymptote at $x = -6$ .

STEP 8

Remember when we divided by $(x - 6)$ ?
That's where a hole might be!

STEP 9

If $x = 6$ , the original function is undefined (because both the numerator and denominator are zero).
But in the simplified version, $x = 6$ gives us $\frac{-4}{6 + 6} = \frac{-4}{12} = -\frac{1}{3}$ .

STEP 10

This means there's a hole at $(6, -\frac{1}{3})$ .

STEP 11

Let's check values close to $x = -6$ .
If $x = -6.1$ , $H(x) = \frac{-4}{-6.1 + 6} = \frac{-4}{-0.1} = 40$ .
If $x = -5.9$ , $H(x) = \frac{-4}{-5.9 + 6} = \frac{-4}{0.1} = -40$ .

STEP 12

So, the graph approaches $\infty$ on one side of $x = -6$ and $-\infty$ on the other.

STEP 13

The degree of the numerator in our simplified function is $0$ (it's just a constant), and the degree of the denominator is $1$ .

STEP 14

When the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y = 0$ .

STEP 15

Vertical asymptote: $x = -6$ .
Hole: $(6, -\frac{1}{3})$ .
The graph approaches $\infty$ on one side of $x = -6$ and $-\infty$ on the other.
Horizontal asymptote: $y = 0$ .

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Studdy Solution

STEP 1

What is this asking? We need to graph $H(x) = \frac{4x - 24}{36 - x^2}$ and answer questions about its vertical asymptotes, holes, and horizontal asymptote. Watch out! Don't forget to simplify the function first!
Holes can be sneaky.

STEP 2

1. Simplify the Function
2. Find Vertical Asymptotes
3. Find Holes
4. Analyze Asymptotic Behavior
5. Find Horizontal Asymptote

STEP 3

We can factor out a $4$ from the numerator: $4x - 24 = 4(x - 6)$ .
So, $H(x) = \frac{4(x - 6)}{36 - x^2}$ .

STEP 4

The denominator is a difference of squares: $36 - x^2 = (6 - x)(6 + x)$ .
Now, $H(x) = \frac{4(x - 6)}{(6 - x)(6 + x)}$ .

STEP 5

STEP 6

Vertical asymptotes happen when the denominator is zero and the numerator isn't also zero at the same $x$ value.
In our simplified function, the denominator is zero when $x + 6 = 0$ , which means $x = -6$ .

STEP 7

So, we have one vertical asymptote at $x = -6$ .

STEP 8

Remember when we divided by $(x - 6)$ ?
That's where a hole might be!

STEP 9

If $x = 6$ , the original function is undefined (because both the numerator and denominator are zero).
But in the simplified version, $x = 6$ gives us $\frac{-4}{6 + 6} = \frac{-4}{12} = -\frac{1}{3}$ .

STEP 10

This means there's a hole at $(6, -\frac{1}{3})$ .

STEP 11

Let's check values close to $x = -6$ .
If $x = -6.1$ , $H(x) = \frac{-4}{-6.1 + 6} = \frac{-4}{-0.1} = 40$ .
If $x = -5.9$ , $H(x) = \frac{-4}{-5.9 + 6} = \frac{-4}{0.1} = -40$ .

STEP 12

So, the graph approaches $\infty$ on one side of $x = -6$ and $-\infty$ on the other.

STEP 13

The degree of the numerator in our simplified function is $0$ (it's just a constant), and the degree of the denominator is $1$ .

STEP 14

When the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y = 0$ .

STEP 15

Vertical asymptote: $x = -6$ .
Hole: $(6, -\frac{1}{3})$ .
The graph approaches $\infty$ on one side of $x = -6$ and $-\infty$ on the other.
Horizontal asymptote: $y = 0$ .

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Studdy Solution

STEP 1

What is this asking? We need to graph H(x)=4x−2436−x2H(x) = \frac{4x - 24}{36 - x^2}H(x)=36−x24x−24​ and answer questions about its vertical asymptotes, holes, and horizontal asymptote. Watch out! Don't forget to simplify the function first!Holes can be sneaky.

STEP 2

1. Simplify the Function2. Find Vertical Asymptotes3. Find Holes4. Analyze Asymptotic Behavior5. Find Horizontal Asymptote

STEP 3

We can **factor out** a 444 from the numerator: 4x−24=4(x−6)4x - 24 = 4(x - 6)4x−24=4(x−6).So, H(x)=4(x−6)36−x2H(x) = \frac{4(x - 6)}{36 - x^2}H(x)=36−x24(x−6)​.

STEP 4

The denominator is a **difference of squares**: 36−x2=(6−x)(6+x)36 - x^2 = (6 - x)(6 + x)36−x2=(6−x)(6+x).Now, H(x)=4(x−6)(6−x)(6+x)H(x) = \frac{4(x - 6)}{(6 - x)(6 + x)}H(x)=(6−x)(6+x)4(x−6)​.

STEP 5

STEP 6

Vertical asymptotes happen when the denominator is **zero** and the numerator isn't also zero at the same xxx value.In our simplified function, the denominator is zero when x+6=0x + 6 = 0x+6=0, which means x=−6x = -6x=−6.

STEP 7

So, we have **one** vertical asymptote at x=−6x = -6x=−6.

STEP 8

Remember when we divided by (x−6)(x - 6)(x−6)?That's where a hole might be!

STEP 9

If x=6x = 6x=6, the original function is undefined (because both the numerator and denominator are zero).But in the simplified version, x=6x = 6x=6 gives us −46+6=−412=−13\frac{-4}{6 + 6} = \frac{-4}{12} = -\frac{1}{3}6+6−4​=12−4​=−31​.

STEP 10

This means there's a **hole** at (6,−13)(6, -\frac{1}{3})(6,−31​).

STEP 11

STEP 12

So, the graph approaches ∞\infty∞ on one side of x=−6x = -6x=−6 and −∞-\infty−∞ on the other.

STEP 13

The **degree** of the numerator in our simplified function is 000 (it's just a constant), and the degree of the denominator is 111.

STEP 14

When the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is y=0y = 0y=0.

STEP 15

Vertical asymptote: x=−6x = -6x=−6.Hole: (6,−13)(6, -\frac{1}{3})(6,−31​).The graph approaches ∞\infty∞ on one side of x=−6x = -6x=−6 and −∞-\infty−∞ on the other.Horizontal asymptote: y=0y = 0y=0.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to graph $H(x) = \frac{4x - 24}{36 - x^2}$ and answer questions about its vertical asymptotes, holes, and horizontal asymptote. Watch out! Don't forget to simplify the function first!
Holes can be sneaky.

1. Simplify the Function
2. Find Vertical Asymptotes
3. Find Holes
4. Analyze Asymptotic Behavior
5. Find Horizontal Asymptote

We can factor out a $4$ from the numerator: $4x - 24 = 4(x - 6)$ .
So, $H(x) = \frac{4(x - 6)}{36 - x^2}$ .

The denominator is a difference of squares: $36 - x^2 = (6 - x)(6 + x)$ .
Now, $H(x) = \frac{4(x - 6)}{(6 - x)(6 + x)}$ .

Vertical asymptotes happen when the denominator is zero and the numerator isn't also zero at the same $x$ value.
In our simplified function, the denominator is zero when $x + 6 = 0$ , which means $x = -6$ .

So, we have one vertical asymptote at $x = -6$ .

Remember when we divided by $(x - 6)$ ?
That's where a hole might be!

If $x = 6$ , the original function is undefined (because both the numerator and denominator are zero).
But in the simplified version, $x = 6$ gives us $\frac{-4}{6 + 6} = \frac{-4}{12} = -\frac{1}{3}$ .

This means there's a hole at $(6, -\frac{1}{3})$ .

So, the graph approaches $\infty$ on one side of $x = -6$ and $-\infty$ on the other.

The degree of the numerator in our simplified function is $0$ (it's just a constant), and the degree of the denominator is $1$ .

When the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y = 0$ .

Vertical asymptote: $x = -6$ .
Hole: $(6, -\frac{1}{3})$ .
The graph approaches $\infty$ on one side of $x = -6$ and $-\infty$ on the other.
Horizontal asymptote: $y = 0$ .