f(x)=x^{2}-4x+9

You can get the vertex form by completing the square.

this is currently in standard form (ax^{2} + bx + c)

In f(x) = x^{2}-4x+9, b = -4

We need to get (b/2)^{2} to complete the square.

b = -4

(b/2)^{2} = (-4/2)^{2 }= (-2)^{2} = 4

Now we take this number and add it to both sides of the equation.

f(x) + 4 = x^{2} - 4x + 9** + 4**

f(x) + 4 = x^{2} - 4x **+ 4** + 9 We rearrange the equation.

f(x) + 4 =** (x ^{2} - 4x + 4)** + 9 Now we can separate the perfect square.

f(x) + 4 = **(x-2) ^{2}** + 9 We can factor the highlighted area.

f(x) = (x-2)^{2} + 9 - 4 Now we solve for f(x)

**f(x) = (x-2) ^{2} + 5 ** This is the vertex form of that equation.

The vertex form is in the format of y = a(x-h)^{2} + k where the vertex is (h, k).

In this case, h = 2 and k = 5 so the vertex is **(2,5)**