Math / Algebra

QuestionGraph the hyperbola using the transverse axis, vertices, and co-vertices: $4 y^{2}-x^{2}-4=0$
Use the green key point to change the orientation of the transverse axis, and the red key points to adjust the locations of the vertices and co-vertices.
Provide your answer below:

Studdy Solution

STEP 1

What is this asking? We need to graph a hyperbola, which is like two mirrored curves, and we'll use some special points to draw it accurately. Watch out! Don't mix up the x and y values!
Also, remember a hyperbola equation has a minus sign between the $x^2$ and $y^2$ terms.

STEP 2

1. Rewrite the equation
2. Find the center
3. Find $a$ and $b$
4. Find the vertices
5. Find the co-vertices
6. Graph the hyperbola

STEP 3

Let's rewrite the equation $4y^2 - x^2 - 4 = 0$ in standard form.
We want it to look like $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ .
First, add 4 to both sides: $4y^2 - x^2 = 4$ .

STEP 4

Now, divide both sides by 4 to get a 1 on the right side: $\frac{4y^2}{4} - \frac{x^2}{4} = \frac{4}{4}$ , which simplifies to $\frac{y^2}{1} - \frac{x^2}{4} = 1$ .
Awesome!

STEP 5

The center of our hyperbola is at $(h, k)$ .
Since we don't have any shifts in our equation, the center is at $(0, 0)$ .
Boom!

STEP 6

From our equation $\frac{y^2}{1} - \frac{x^2}{4} = 1$ , we see that $a^2 = 1$ and $b^2 = 4$ .
Taking the square root of both sides gives us $a = 1$ and $b = 2$ .
These values will help us find the vertices and co-vertices!

STEP 7

Since the $y^2$ term is positive, the transverse axis is vertical.
The vertices are $a$ units above and below the center.
So, our vertices are at $(0, 1)$ and $(0, -1)$ .

STEP 8

The co-vertices are $b$ units to the left and right of the center.
Since $b = 2$ , our co-vertices are at $(2, 0)$ and $(-2, 0)$ .

STEP 9

Now, plot the center, vertices, and co-vertices on the graph.

STEP 10

Draw a rectangle using the vertices and co-vertices.
The asymptotes are the lines that pass through the diagonals of this rectangle.

STEP 11

Finally, sketch the hyperbola curves, making sure they approach the asymptotes but never touch them.
The curves pass through the vertices and open upwards and downwards because the $y^2$ term is positive.

STEP 12

The hyperbola is centered at $(0, 0)$ , has vertices at $(0, 1)$ and $(0, -1)$ , and co-vertices at $(2, 0)$ and $(-2, 0)$ .
The transverse axis is vertical.

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QuestionGraph the hyperbola using the transverse axis, vertices, and co-vertices: $4 y^{2}-x^{2}-4=0$
Use the green key point to change the orientation of the transverse axis, and the red key points to adjust the locations of the vertices and co-vertices.
Provide your answer below:

Studdy Solution

STEP 1

What is this asking? We need to graph a hyperbola, which is like two mirrored curves, and we'll use some special points to draw it accurately. Watch out! Don't mix up the x and y values!
Also, remember a hyperbola equation has a minus sign between the $x^2$ and $y^2$ terms.

STEP 2

1. Rewrite the equation
2. Find the center
3. Find $a$ and $b$
4. Find the vertices
5. Find the co-vertices
6. Graph the hyperbola

STEP 3

Let's rewrite the equation $4y^2 - x^2 - 4 = 0$ in standard form.
We want it to look like $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ .
First, add 4 to both sides: $4y^2 - x^2 = 4$ .

STEP 4

Now, divide both sides by 4 to get a 1 on the right side: $\frac{4y^2}{4} - \frac{x^2}{4} = \frac{4}{4}$ , which simplifies to $\frac{y^2}{1} - \frac{x^2}{4} = 1$ .
Awesome!

STEP 5

The center of our hyperbola is at $(h, k)$ .
Since we don't have any shifts in our equation, the center is at $(0, 0)$ .
Boom!

STEP 6

From our equation $\frac{y^2}{1} - \frac{x^2}{4} = 1$ , we see that $a^2 = 1$ and $b^2 = 4$ .
Taking the square root of both sides gives us $a = 1$ and $b = 2$ .
These values will help us find the vertices and co-vertices!

STEP 7

Since the $y^2$ term is positive, the transverse axis is vertical.
The vertices are $a$ units above and below the center.
So, our vertices are at $(0, 1)$ and $(0, -1)$ .

STEP 8

The co-vertices are $b$ units to the left and right of the center.
Since $b = 2$ , our co-vertices are at $(2, 0)$ and $(-2, 0)$ .

STEP 9

Now, plot the center, vertices, and co-vertices on the graph.

STEP 10

Draw a rectangle using the vertices and co-vertices.
The asymptotes are the lines that pass through the diagonals of this rectangle.

STEP 11

Finally, sketch the hyperbola curves, making sure they approach the asymptotes but never touch them.
The curves pass through the vertices and open upwards and downwards because the $y^2$ term is positive.

STEP 12

The hyperbola is centered at $(0, 0)$ , has vertices at $(0, 1)$ and $(0, -1)$ , and co-vertices at $(2, 0)$ and $(-2, 0)$ .
The transverse axis is vertical.

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Studdy Solution

STEP 1

What is this asking? We need to graph a hyperbola, which is like two mirrored curves, and we'll use some special points to draw it accurately. Watch out! Don't mix up the *x* and *y* values!Also, remember a hyperbola equation has a minus sign between the x2x^2x2 and y2y^2y2 terms.

STEP 2

1. Rewrite the equation2. Find the center3. Find aaa and bbb4. Find the vertices5. Find the co-vertices6. Graph the hyperbola

STEP 3

Let's **rewrite** the equation 4y2−x2−4=04y^2 - x^2 - 4 = 04y2−x2−4=0 in standard form.We want it to look like y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1a2y2​−b2x2​=1.First, **add** 4 to both sides: 4y2−x2=44y^2 - x^2 = 44y2−x2=4.

STEP 4

Now, **divide** both sides by 4 to get a 1 on the right side: 4y24−x24=44\frac{4y^2}{4} - \frac{x^2}{4} = \frac{4}{4}44y2​−4x2​=44​, which simplifies to y21−x24=1\frac{y^2}{1} - \frac{x^2}{4} = 11y2​−4x2​=1.Awesome!

STEP 5

The center of our hyperbola is at (h,k)(h, k)(h,k).Since we don't have any shifts in our equation, the **center** is at (0,0)(0, 0)(0,0).Boom!

STEP 6

From our equation y21−x24=1\frac{y^2}{1} - \frac{x^2}{4} = 11y2​−4x2​=1, we see that a2=1a^2 = 1a2=1 and b2=4b^2 = 4b2=4.Taking the **square root** of both sides gives us a=1a = 1a=1 and b=2b = 2b=2.These values will help us find the vertices and co-vertices!

STEP 7

Since the y2y^2y2 term is positive, the **transverse axis** is vertical.The vertices are aaa units above and below the center.So, our **vertices** are at (0,1)(0, 1)(0,1) and (0,−1)(0, -1)(0,−1).

STEP 8

The co-vertices are bbb units to the left and right of the center.Since b=2b = 2b=2, our **co-vertices** are at (2,0)(2, 0)(2,0) and (−2,0)(-2, 0)(−2,0).

STEP 9

Now, **plot** the center, vertices, and co-vertices on the graph.

STEP 10

Draw a rectangle using the vertices and co-vertices.The **asymptotes** are the lines that pass through the diagonals of this rectangle.

STEP 11

Finally, **sketch** the hyperbola curves, making sure they approach the asymptotes but never touch them.The curves pass through the vertices and open upwards and downwards because the y2y^2y2 term is positive.

STEP 12

The hyperbola is centered at (0,0)(0, 0)(0,0), has vertices at (0,1)(0, 1)(0,1) and (0,−1)(0, -1)(0,−1), and co-vertices at (2,0)(2, 0)(2,0) and (−2,0)(-2, 0)(−2,0).The transverse axis is vertical.

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What is this asking? We need to graph a hyperbola, which is like two mirrored curves, and we'll use some special points to draw it accurately. Watch out! Don't mix up the x and y values!
Also, remember a hyperbola equation has a minus sign between the $x^2$ and $y^2$ terms.

1. Rewrite the equation
2. Find the center
3. Find $a$ and $b$
4. Find the vertices
5. Find the co-vertices
6. Graph the hyperbola

Let's rewrite the equation $4y^2 - x^2 - 4 = 0$ in standard form.
We want it to look like $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ .
First, add 4 to both sides: $4y^2 - x^2 = 4$ .

Now, divide both sides by 4 to get a 1 on the right side: $\frac{4y^2}{4} - \frac{x^2}{4} = \frac{4}{4}$ , which simplifies to $\frac{y^2}{1} - \frac{x^2}{4} = 1$ .
Awesome!

The center of our hyperbola is at $(h, k)$ .
Since we don't have any shifts in our equation, the center is at $(0, 0)$ .
Boom!

From our equation $\frac{y^2}{1} - \frac{x^2}{4} = 1$ , we see that $a^2 = 1$ and $b^2 = 4$ .
Taking the square root of both sides gives us $a = 1$ and $b = 2$ .
These values will help us find the vertices and co-vertices!

Since the $y^2$ term is positive, the transverse axis is vertical.
The vertices are $a$ units above and below the center.
So, our vertices are at $(0, 1)$ and $(0, -1)$ .

The co-vertices are $b$ units to the left and right of the center.
Since $b = 2$ , our co-vertices are at $(2, 0)$ and $(-2, 0)$ .

Now, plot the center, vertices, and co-vertices on the graph.

Draw a rectangle using the vertices and co-vertices.
The asymptotes are the lines that pass through the diagonals of this rectangle.

Finally, sketch the hyperbola curves, making sure they approach the asymptotes but never touch them.
The curves pass through the vertices and open upwards and downwards because the $y^2$ term is positive.

The hyperbola is centered at $(0, 0)$ , has vertices at $(0, 1)$ and $(0, -1)$ , and co-vertices at $(2, 0)$ and $(-2, 0)$ .
The transverse axis is vertical.