Find the equation of the parabola whose axis is horizontal, vertex on the y-axis and which passes through (2,4) and (8,-2).

the general formula for a parabola with known vertex and horizontal axis is

x = a(y-k)^2 + h, where (h,k) is the vertex

in your case the vertex , being on the y-axis , can be called (0,k) , and the equation would be
x = a(y - k)^2 + 0
for the point (2,4)
2 = a(4-k)^2 (#1)

for the point (8,-2)
8 = a(-2-k)^2 (#2)

divide #2 by #1
4 = (-2-k)^2 / (4-k)^2
4(4-k)^2 = (-2-k)^2
64 - 32k + 4k^2 = 4 + 4k + k^2
3k^2 -36 + 60 = 0
k^2 - 12 + 20 = 0
(k-10)(k-2) = 0
k = 10 or k = 2

if k=2, x = a(y-2)^2
and using (2,4)
2 = a(4-2)^2
a = 1/2 ---------> x = (1/2)(y-2)^2

if k=10 , x = a(y-10)^2
again using (2,4)
2 = a(4-10)^2
a = 1/18 ----------> x= (1/18)(y-10)^2

Notice that there are two possible equations,
btw, I checked both equations using the point(8,-2), it also satisfies both equations.

Why did the parabola see a therapist? Because it had trouble finding its vertex on the y-axis! But no worries, I'm here to help you out!

First, let's consider the general equation of a parabola with a horizontal axis, given by y = a(x - h)^2 + k. The vertex of this parabola would be (h, k), where h represents the x-coordinate and k represents the y-coordinate.

Since the parabola's vertex is on the y-axis, we know that h = 0 and k = 0. This simplifies the equation to y = ax^2.

Now, let's use the given points (2,4) and (8,-2) to find the value of a.

Substituting the coordinates of (2,4) into the equation, we get:
4 = a(2)^2
4 = 4a
a = 1

Substituting the coordinates of (8,-2) into the equation, we get:
-2 = a(8)^2
-2 = 64a
a = -1/32

Wait, something doesn't seem right! It appears that the values of a obtained from both points are different. This means that there is no single parabola that passes through both points and has its vertex on the y-axis.

I apologize if I couldn't provide the equation you were looking for. But hey, at least we had a little fun with the parabola's therapy session, right?

To find the equation of the parabola, we need to determine its standard form, which is given by the equation:

y = a(x - h)^2 + k

where (h, k) represents the coordinates of the vertex.

Since the axis of the parabola is horizontal and the vertex lies on the y-axis, the equation can be simplified as:

y = a(x - 0)^2 + k
y = ax^2 + k

Now we need to find the values of a and k.

Given that the parabola passes through (2,4), we can substitute these coordinates into the equation to get:

4 = a(2^2) + k
4 = 4a + k -- Eq. 1

Similarly, using the point (8,-2):

-2 = a(8^2) + k
-2 = 64a + k -- Eq. 2

Now we have a system of equations with two variables (a and k):
4 = 4a + k -- Eq. 1
-2 = 64a + k -- Eq. 2

To solve this system, we'll subtract Eq. 2 from Eq. 1:

4 - (-2) = (4a + k) - (64a + k)

Simplifying this equation:

6 = 4a - 64a

Combining like terms:

6 = -60a

Dividing both sides by -60:

a = -1/10

Now that we have the value of a, we can substitute it back into Eq. 1 or Eq. 2 to solve for k. Let's use Eq. 1:

4 = 4(-1/10) + k
4 = -4/10 + k

Combining like terms:

4 = -2/5 + k

To get rid of the fraction, we can rewrite 4 as 20/5:

20/5 = -2/5 + k

Now we have a common denominator:

20/5 = (-2 + 5k) / 5

Multiplying both sides by 5:

20 = -2 + 5k

Adding 2 to both sides:

20 + 2 = -2 + 2 + 5k

Simplifying:

22 = 5k

Dividing both sides by 5:

k = 22/5

Therefore, the equation of the parabola is:

y = (-1/10)x^2 + 22/5

To find the equation of a parabola, we need to use the general form of the equation for a parabola with a horizontal axis:

y = a(x - h)^2 + k

Here, (h, k) is the vertex of the parabola. Since the vertex of the parabola is on the y-axis, the x-coordinate of the vertex is 0. Therefore, we have h = 0.

Let's first find the value of k, the y-coordinate of the vertex. We know that the vertex lies on the y-axis, so it has the form (0, k). We need to find k.

Now we are given two points that lie on the parabola: (2, 4) and (8, -2). We can substitute these coordinates to find the values of k and a.

For the point (2,4), substituting x = 2 and y = 4 into the equation y = a(x - h)^2 + k, we get:

4 = a(2 - 0)^2 + k
4 = 4a + k ----(1)

Similarly, for the point (8,-2), substituting x = 8 and y = -2, we get:

-2 = a(8 - 0)^2 + k
-2 = 64a + k ----(2)

Now we have a system of two equations with two variables: a and k. We can solve this system to find the values of a and k.

Subtracting equation (1) from equation (2), we get:

-2 - 4 = 64a + k - (4a + k)
-6 = 60a
a = -1/10 ----(3)

Substituting this value of a back into equation (1), we can solve for k:

4 = 4(-1/10) + k
4 = -4/10 + k
4 = -2/5 + k
4 + 2/5 = k
k = 22/5 ----(4)

Therefore, the equation of the parabola is:

y = (-1/10)x^2 + 22/5