How do I find vertex, focus and directrix for the following functions?

y^2−16y=12x−8^2

x^2+16x=4y−24

(x−2)^2=12(y−2)

You just need to remember the characteristics of parabolas.

y^2?16y=12x?8^2
y^2-16y+64 = 12x-64+64
(y-8)^2 = 12x

Now recall that for y^2 = 4px, we have
vertex = (0,0)
focus = (p,0)
directrix is x = -p

So, for your parabola, that means that it is shifted up 8, and p=3:

vertex = (0,8)
focus = (3,8)
directrix is x = -3

http://www.wolframalpha.com/input/?i=parabola+y%5E2%E2%88%9216y%3D12x%E2%88%928%5E2

Now massage the other equations till they fit the general formula for

(x-h)^2 = 4p(y-k)
or
(y-k)^2 = 4p(x-h)

and then apply the required shifts

Well, well, well! It looks like you're looking for the vertex, focus, and directrix of these quadratic functions. Let's get this show on the road!

For the first function, y^2 − 16y = 12x − 8^2, we need to rewrite the equation in the standard form of a parabola. Completing the square for the y's, we get:
(y - 8)^2 = 12(x - 16)
Now we can easily identify the vertex as (16, 8). The parabola opens to the right, so you won't find a directrix. But to find the focus, we need to know the value of p, which is the distance from the vertex to the focus. In this case, p = 1/4a = 1/4(12) = 3. This means the focus is at (16 + 3, 8) = (19, 8).

For the second function, x^2 + 16x = 4y − 24, we again need to rewrite it in the standard form. Completing the square for the x's, we have:
(x + 8)^2 = 4(y - 6)
Oh look, the vertex just unveiled itself as (-8, 6). Now, the parabola opens upwards, so the directrix lies below the vertex at y = -2. To find the focus, we need to know p. In this case, p = 1/4a = 1/4(4) = 1. So the focus is at (-8, 6 + 1) = (-8, 7).

Lastly, for (x - 2)^2 = 12(y - 2), the equation is already in a splendid form! The vertex, as you can spot, is (2, 2). Now, since it opens upwards, the directrix is found below the vertex at y = 1. To discover the focus, we need p, which is 1/4a = 1/4(12) = 3. Thus, the focus is (2, 2 + 3) = (2, 5).

Remember, laughter is the best vertex, focus, and directrix finder!

1) Finding vertex, focus, and directrix for the function: y^2 - 16y = 12x - 8^2

Step 1: Rewrite the equation in standard form by completing the square. In this case, we need to complete the square for both x and y.

y^2 - 16y + 64 = 12x
(y - 8)^2 = 12x

Step 2: Compare the equation to the standard form of a parabola: (y - k)^2 = 4p(x - h), where (h, k) is the vertex.

In our case, the vertex is at (h, k) = (0, 8).

Step 3: Find the value of p using the equation 4p = 12. Therefore, p = 3.

Step 4: The focus (F) and directrix (D) can now be determined. The focus is at (h + p, k), so the coordinates are (3, 8). The directrix is a horizontal line y = k - p, which in this case, is y = 8 - 3.

So, for the function y^2 - 16y = 12x - 8^2, the vertex is (0, 8), the focus is (3, 8), and the directrix is y = 5.

2) Finding vertex, focus, and directrix for the function: x^2 + 16x = 4y - 24

Step 1: Rewrite the equation in standard form by completing the square. In this case, we need to complete the square for both x and y.

x^2 + 16x + 64 = 4y
(x + 8)^2 = 4y + 64

Step 2: Compare the equation to the standard form of a parabola: (x - h)^2 = 4p(y - k), where (h, k) is the vertex.

In our case, the vertex is at (h, k) = (-8, 0).

Step 3: Find the value of p using the equation 4p = 4. Therefore, p = 1.

Step 4: The focus (F) and directrix (D) can now be determined. The focus is at (h, k + p), so the coordinates are (-8, 1). The directrix is a vertical line x = h - p, which in this case, is x = -9.

So, for the function x^2 + 16x = 4y - 24, the vertex is (-8, 0), the focus is (-8, 1), and the directrix is x = -9.

3) Finding vertex, focus, and directrix for the function: (x - 2)^2 = 12(y - 2)

Step 1: Rewrite the equation in standard form by expanding the equation on the left side.

x^2 - 4x + 4 = 12y - 24

Step 2: Compare the equation to the standard form of a parabola: (x - h)^2 = 4p(y - k), where (h, k) is the vertex.

In our case, the vertex is at (h, k) = (2, 0).

Step 3: Find the value of p using the equation 4p = 12. Therefore, p = 3.

Step 4: The focus (F) and directrix (D) can now be determined. The focus is at (h, k + p), so the coordinates are (2, 3). The directrix is a horizontal line y = k - p, which in this case, is y = -3.

So, for the function (x - 2)^2 = 12(y - 2), the vertex is (2, 0), the focus is (2, 3), and the directrix is y = -3.

To find the vertex, focus, and directrix for the given equations, we can compare them to the standard form equations for parabolas.

1. For the equation: y^2 - 16y = 12x - 8^2

To find the vertex, we complete the square for both x and y. Let's focus on y first:

y^2 - 16y = 12x - 8^2

Rearranging the terms:

(y^2 - 16y + 64) = 12x

Completing the square for y:

(y^2 - 16y + 64) = 12x + 0

(y - 8)^2 = 12x

Now we can identify the vertex from the standard form, which is (h, k). In this case, h = 0 and k = 8. Therefore, the vertex is (0, 8).

To find the focus and directrix, we need to determine whether the parabola opens horizontally or vertically. Since the equation is in the form (y - k)^2 = 4a(x - h), where a is the distance from the vertex to the focus and from the vertex to the directrix, we can see that a = 3 (since 2a = 12, so a = 6) and the parabola opens horizontally.

The focus is located at (h + a, k), so the focus is (3, 8).

The directrix is a horizontal line that is a distance of a units away from the vertex. In this case, the directrix is k + a units away from the vertex, so the directrix is the line y = 8.

Therefore, for the equation y^2 - 16y = 12x - 8^2, the vertex is (0, 8), the focus is (3, 8), and the directrix is y = 8.

2. For the equation: x^2 + 16x = 4y - 24

The process is similar to the previous equation:

x^2 + 16x = 4y - 24

Rearranging the terms:

(x^2 + 16x + 64) = 4y

(x + 8)^2 = 4y + 64

Identifying the vertex, h = -8 and k = -16. Therefore, the vertex is (-8, -16).

Since the equation is in the form (x - h)^2 = 4a(y - k), we can identify that a = 4 (since 2a = 16, so a = 8) and the parabola opens vertically.

The focus is located at (h, k + a), so the focus is (-8, -16 + 4) which simplifies to (-8, -12).

The directrix is a vertical line that is a distance of a units away from the vertex. In this case, the directrix is x = -8.

Therefore, for the equation x^2 + 16x = 4y - 24, the vertex is (-8, -16), the focus is (-8, -12), and the directrix is x = -8.

3. For the equation: (x - 2)^2 = 12(y - 2)

Using a similar approach:

(x - 2)^2 = 12(y - 2)

Identifying the vertex, h = 2 and k = 2. Therefore, the vertex is (2, 2).

Since the equation is in the form (x - h)^2 = 4a(y - k), we can deduce that a = 3.

The focus is located at (h, k + a), so the focus is (2, 2 + 3) which simplifies to (2, 5).

The directrix is a horizontal line that is a distance of a units away from the vertex. In this case, the directrix is y = 2 - 3 which simplifies to y = -1.

Therefore, for the equation (x - 2)^2 = 12(y - 2), the vertex is (2, 2), the focus is (2, 5), and the directrix is y = -1.

These are the vertex, focus, and directrix for the given equations.