What is the y-coordinate of the vertex of a parabola with the following equation y equals x2 - 8x plus 18?

You can work this out by taking the derivative of the equation, and solving for zero:

y = x2 - 8x + 18

y' = 2x - 8

0 = 2x - 8

x = 4

So the vertex occurs where x is equal to 4. You can then plug that back into the original equation to get the y-coordinate:

y = 42 - 8(4) + 18

y = 16 - 32 + 18

y = 2

So the vertex of the parabola occurs at the point (4, 2), leaving 2 as the answer to your question.