find the vertex and line of symmetry

f(x)=x^2-6x+8
f(x)=3x^2-18x+6
f(x)=4x^2-8x+
f(x)=-2(x+6)^2+6

f(x)=x^2-6x+8=2x-6

f(x)=3x^2-18x+6=6x-18

f(x)=4x^2-8x+=8x-8

To find the vertex and line of symmetry of a quadratic function in the form f(x) = ax^2 + bx + c, you can use the formula:

Vertex x-coordinate = -b / (2a)
Vertex y-coordinate = f(vertex x-coordinate)

Let's find the vertex and line of symmetry for each of the given quadratic functions:

1. f(x) = x^2 - 6x + 8
Here, a = 1, b = -6, and c = 8.
Vertex x-coordinate = -(-6) / (2 * 1) = 6 / 2 = 3
To find the y-coordinate, substitute x = 3 into the equation:
f(3) = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1
So, the vertex is (3, -1).
The line of symmetry is x = 3.

2. f(x) = 3x^2 - 18x + 6
Here, a = 3, b = -18, and c = 6.
Vertex x-coordinate = -(-18) / (2 * 3) = 18 / 6 = 3
To find the y-coordinate, substitute x = 3 into the equation:
f(3) = 3(3)^2 - 18(3) + 6 = 27 - 54 + 6 = -21
So, the vertex is (3, -21).
The line of symmetry is x = 3.

3. f(x) = 4x^2 - 8x (missing c value)
Here, a = 4, b = -8.
Vertex x-coordinate = -(-8) / (2 * 4) = 8 / 8 = 1
To find the y-coordinate, substitute x = 1 into the equation:
f(1) = 4(1)^2 - 8(1) = 4 - 8 = -4
So, the vertex is (1, -4).
The line of symmetry is x = 1.

4. f(x) = -2(x+6)^2 + 6
Here, a = -2, b = -12, and c = 6. (note that b = -2 * 6 = -12)
Vertex x-coordinate = -(-12) / (2 * -2) = 12 / -4 = -3
To find the y-coordinate, substitute x = -3 into the equation:
f(-3) = -2((-3)+6)^2 + 6 = -2(3)^2 + 6 = -2(9) + 6 = -12
So, the vertex is (-3, -12).
The line of symmetry is x = -3.

Therefore, for the given quadratic functions:
1. f(x) = x^2 - 6x + 8
Vertex = (3, -1)
Line of symmetry: x = 3

2. f(x) = 3x^2 - 18x + 6
Vertex = (3, -21)
Line of symmetry: x = 3

3. f(x) = 4x^2 - 8x
Vertex = (1, -4)
Line of symmetry: x = 1

4. f(x) = -2(x+6)^2 + 6
Vertex = (-3, -12)
Line of symmetry: x = -3