how do i find the equation of a parabola given x intercepts x=0 and x=6 and a vertex at (3,4)?

you know the equation must be of the form

y = a(x-3)^2 + 4
all we need is the value of a to stear it so it passes through (6,0)
0 = a(6-3)^2+4
0 = 9a + 4
a = -4/9

so: y = (-4/9)(x-3)^2 + 4

*facepalm* wow i can't believe i forgot that. thanks for including the

y=a(x-3)^2+4 i cant believe i forgot to use point-slope form.

To find the equation of a parabola given the x-intercepts and the vertex, we can start by using the general form of a parabolic equation:

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

where (h, k) is the vertex of the parabola, and a is a constant that determines the shape of the parabola.

Given the x-intercepts x = 0 and x = 6, we know that the y-coordinate at both of these points is 0.

So let's substitute those values into the equation to get two equations:

0 = a(0-h)^2 + k —— (1)
0 = a(6-h)^2 + k —— (2)

We also know that the vertex is (3, 4), so we can substitute those values into the equation to get one more equation:

4 = a(3-h)^2 + k —— (3)

Now we have a system of three equations (equations 1, 2, and 3) with three unknowns (a, h, and k). We can solve this system to find the values of a, h, and k.

First, let's simplify equation (1):

0 = ah^2 + k —— (1')

Now subtract equation (1') from equation (2):

0 = a(6-h)^2 - (ah^2 + k)
0 = 36a - ah^2 - k —— (4)

Next, let's simplify equation (3):

4 = 9a - 6ah + ah^2 + k —— (5)

Now we have equations (4) and (5) as a system of two equations with two unknowns (a and h). We can solve these equations simultaneously.

To eliminate k, let's multiply equation (4) by 1:

0 = 36a - ah^2 - k —— (4')
0 = 36a - ah^2 —— (4'')

Now let's subtract equation (5) from equation (4''):

0 = 36a - ah^2 - (9a - 6ah + ah^2 + k)
0 = 36a - 9a - ah^2 + 6ah - ah^2 + k
0 = 27a - 2ah^2 + 6ah + k —— (6)

Now we have one equation (6) with two unknowns (a and h). Since we know the vertex is (3, 4), we substitute h = 3 and k = 4 into equation (6):

0 = 27a - 2(3^2) + 6(3) + 4
0 = 27a - 2(9) + 6(3) + 4
0 = 27a - 18 + 18 + 4
0 = 27a + 4

To solve for a, we now isolate it:

27a = -4

Dividing both sides by 27:

a = -4/27

Now we know the value of a. Let's substitute it back into equation (1') to solve for k:

0 = (-4/27)h^2 + k

Substituting h = 3 and solving for k:

0 = (-4/27)(3^2) + k
0 = (-4/27)(9) + k
0 = -36/27 + k
0 = -4/3 + k

To solve for k, isolate it:

4/3 = k

Now we have the value of k.

Putting all the values together, the equation of the parabola is:

y = (-4/27)(x-3)^2 + 4/3

To find the equation of a parabola, you need to use the general form of a quadratic equation: y = ax^2 + bx + c, where a, b, and c are constants to be determined.

Given that the parabola has x-intercepts at x = 0 and x = 6, we can determine the two points where the parabola intersects the x-axis. These points are (0, 0) and (6, 0).

Since the vertex of the parabola is given at (3, 4), we know that the x-coordinate of the vertex, 3, lies on the axis of symmetry. The axis of symmetry is equidistant from the two x-intercepts, so it passes through the midpoint of the two intercepts. Thus, the midpoint of (0, 0) and (6, 0) is (3, 0).

Now we have three points: the vertex (3, 4), and the x-intercepts (0, 0) and (6, 0). We can use these points to determine the values of a, b, and c.

Step 1: Find the value of a:
Since the parabola is in the form y = ax^2 + bx + c, we can substitute the x and y coordinates of the vertex (3, 4) into the equation:
4 = a * 3^2 + b * 3 + c
16 = 9a + 3b + c

Step 2: Find the value of b:
Substitute the x and y coordinates of the x-intercept (0, 0) into the equation:
0 = a * 0^2 + b * 0 + c
0 = c

Step 3: Find the value of b (continued):
Substitute the x and y coordinates of the x-intercept (6, 0) into the equation:
0 = a * 6^2 + b * 6 + c
0 = 36a + 6b

Now we have a system of equations:
16 = 9a + 3b
0 = 36a + 6b

Step 4: Solve the system of equations:
Solve the system of equations simultaneously to find the values of a and b:
From the second equation, we can rewrite it as:
0 = 6b - 36a
6b = 36a
b = 6a

Substitute the value of b in terms of a into the first equation:
16 = 9a + 3(6a)
16 = 9a + 18a
16 = 27a
a = 16/27

Substitute the value of a back into b = 6a:
b = 6 * (16/27)
b = 96/27
b = 32/9

Since c = 0 from Step 2, we now have the values of a, b, and c:
a = 16/27
b = 32/9
c = 0

Therefore, the equation of the parabola is:
y = (16/27)x^2 + (32/9)x + 0
Simplifying,
y = (16/27)x^2 + (32/9)x