f(x) is a quadratic function such that f(1) = 3 and f(5) = 3. Find the x coordinate of the vertex of the graph of f.

well, since f(1) = f(5), the vertex is at x=3, midway between. So,

y = a(x-3)^2 + k
Now plug in your points to get a and k:

3-k = 4a
k = 3-4a

y = a(x-3)^2 + 3-4a

Unfortunately, since all we have is two points, there's no way to determine exactly where the vertex is. It is somewhere on the line x=3.

However, if we take a=1, then we have

y = (x-3)^2 - 1
y+1 = (x-3)^2
and the vertex is at (3,-1)

To find the x-coordinate of the vertex of the graph of f(x), we need to use the formula for the x-coordinate of the vertex of a quadratic function, which is given by x = -b/2a, where the quadratic function is in the form f(x) = ax^2 + bx + c.

In this case, we have two points on the graph of f(x) that we can use to find the values of a, b, and c.

Given that f(1) = 3 and f(5) = 3, we can set up two equations using these points:

1) f(1) = a(1)^2 + b(1) + c = 3
a + b + c = 3

2) f(5) = a(5)^2 + b(5) + c = 3
25a + 5b + c = 3

Now, we have a system of equations that can be solved simultaneously. Subtracting equation 1) from equation 2), we get:

25a + 5b + c - (a + b + c) = 3 - 3
24a + 4b = 0

Dividing both sides of this equation by 4, we have:

6a + b = 0

Now, we have a system of two equations with two variables:

a + b + c = 3
6a + b = 0

To solve this system, we can substitute the value of b from the second equation into the first equation:

a + (-6a) + c = 3
-5a + c = 3

Now, we have a single equation with one variable that we can solve for a:

-5a + c = 3
-5a = 3 - c
a = (3 - c)/(-5)

Since the x-coordinate of the vertex is given by x = -b/2a, we need to find the value of b. We can substitute the values of a and c into the equation 6a + b = 0 to solve for b:

6a + b = 0
6((3 - c)/(-5)) + b = 0
(18 - 6c)/(-5) + b = 0
b = (6c - 18)/5

Now, we can substitute the values of a and b into the equation x = -b/2a to find the x-coordinate of the vertex:

x = -((6c - 18)/5)/(2((3 - c)/(-5)))
x = -((6c - 18)/5)*(5/(2(3 - c)))
x = -(6c - 18)/(2(3 - c))
x = -(6c - 18)/(6 - 2c)
x = -(2(3c - 9))/(6 - 2c)
x = (2(9 - 3c))/(2(3 - c))
x = (9 - 3c)/(3 - c)

Therefore, the x-coordinate of the vertex of the graph of f is given by (9 - 3c)/(3 - c).

To find the x-coordinate of the vertex of a quadratic function, we can use the formula:

x = -b / (2a)

where a, b, and c are the coefficients of the quadratic function in the form f(x) = ax^2 + bx + c.

In this case, we don't have the equation of the quadratic function, but we are given some points on its graph. We can use this information to determine the values of a, b, and c, and then find the x-coordinate of the vertex.

We are given that f(1) = 3 and f(5) = 3. This means we have two points on the graph of the quadratic function: (1, 3) and (5, 3).

Let's substitute these points into the general form of a quadratic function to find the coefficients.

For point (1, 3):
3 = a(1)^2 + b(1) + c
3 = a + b + c ... Equation 1

For point (5, 3):
3 = a(5)^2 + b(5) + c
3 = 25a + 5b + c ... Equation 2

We now have a system of two equations (Equation 1 and Equation 2) with three variables (a, b, and c). We need an additional equation to solve for the three variables.

One way to find this additional equation is by considering the x-coordinate of the vertex. Since a parabola is symmetric with respect to its vertex, the x-coordinate of the vertex is the average of the x-coordinates of the given points.

x-coordinate of the vertex = (1 + 5) / 2 = 6 / 2 = 3

So, we know that the x-coordinate of the vertex is 3.

Now, substitute this x-coordinate into the quadratic function to solve for a, b, and c.

Using x = 3 in Equation 1:
3 = a(3)^2 + b(3) + c
3 = 9a + 3b + c

Using x = 3 in Equation 2:
3 = 25a + 5b + c

Now, we have a system of two equations and three variables. We can solve this system to find the values of a, b, and c.

Subtracting the second equation from the first equation, we have:
0 = -16a - 2b

Simplifying, we get:
-16a - 2b = 0 ... Equation 3

Now, substitute the value of b from Equation 3 into one of the original equations (either Equation 1 or Equation 2) to solve for a. Let's substitute into Equation 1:

3 = a + b + c

Substituting b = (-16a) / 2 from Equation 3:
3 = a + (-16a) / 2 + c
3 = (2a - 16a) / 2 + c
3 = -14a / 2 + c
3 = -7a + c ... Equation 4

Now, we have two equations with two variables (Equation 3 and Equation 4). We can solve this system to find the values of a and c.

Multiplying Equation 3 by 7, we have:
0 = -112a - 14b ... Equation 5

Adding Equation 5 and Equation 4, we have:
3 = -112a - 14a + c
3 = -126a + c

Now, we can isolate c:
c = 3 + 126a

Substituting this into Equation 1, we have:
3 = a + b + (3 + 126a)
3 = a + b + 3 + 126a
0 = 127a + b

We now have a system of two equations and two variables to solve:
0 = 127a + b ... Equation 6
c = 3 + 126a

To find the values of a and b, we can substitute Equation 6 into Equation 5:
0 = -112a - 14(127a)
0 = -112a - 1778a
0 = -1890a

To solve for a, we can divide both sides by -1890:
a = 0

Now, substitute a = 0 into Equation 6:
0 = 127(0) + b
0 = b

Finally, substitute a = 0 and b = 0 into Equation 4 to find c:
3 = -7(0) + c
3 = c

Now we have the values of a, b, and c:
a = 0, b = 0, c = 3

Plugging these values back into the general form of a quadratic function:
f(x) = ax^2 + bx + c
f(x) = 0x^2 + 0x + 3
f(x) = 3

So, the quadratic function is f(x) = 3, which means it is a horizontal line passing through the point (3, 3).

Therefore, the x-coordinate of the vertex of the graph of f is 3.

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