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# What are the coordinates of the vertex for f(x) = x^{2} + 6x + 13?

**Solution:**

We have to find the coordinates of the vertex.

For any equation of the type y = ax^{2} + bx + c, the vertex is given by (h,k)

Where, h = -b/2a and k = (4ac - b^{2})/4a.

Given, the equation is y = x^{2} + 6x + 13

Here, a = 1, b = 6 and c = 13

So, h = -6/2(1)

h = -6/2

h = -3

4ac = 4(1)(13)

4ac = 52

So, k = [52 - (6)^{2}]/4(1)

k = (52 - 36)/4

k = 16/4

k = 4

Therefore, the coordinates of the vertex is (h, k) = (-3, 4).

## What are the coordinates of the vertex for f(x) = x^{2} + 6x + 13?

**Summary:**

The coordinates of the vertex for f(x) = x^{2} + 6x + 13 is (-3, 4).

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