what is the vertex of x^2+8x+5

y = x^2+8x+5

In the form y = ax^2 + bx + c, the x coordinate of the vertex is given by x = -b/2a

So here, a = 1 and b = 8

So  x = -8/2(1) = -4

And to find the y coordinate of the vertex, we just put the value for x back into the function.....so we have (-4)^2 + 8(-4) + 5 = 16 - 32 + 5 = -16 + 5 = -11

So the vertex is (-4, -11).....

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To answer the other two questions

x^2 + 18x + _____

Just take 1/2 of the coefficient on the x variable and square the resulting value.  Thus (1/2)18 = 9.....and 9^2 = 81. Thus, 81  is the number needed to complete the square (ensure a "perfect square trinomial)

For the second one, we have (1/2)(-5) = (-5/2)   and (-5/2)^2  = 25/4