Question

Find the vertex of the parabola whose equation is y = -2x2 + 8x - 5.

Answer 1

The vertex is the point where the graph is horizontal, which you can find by setting the first derivative to zero.

derivative: -4x + 8 = 0 => x=2
fill it in in the equation:

y=-2*2² + 8*2 - 5 = 3

So the answer is the point (2,3)

Answer 2

In order to do this you have to put it into vertex form, which is achieved by completing the square.  Start out by setting it equal to 0. $-2 x^{2} +8x-5=0$.  Next, move the 5 over to the other side of the equals sign. $-2 x^{2} +8x=5$.  In order to complete the square, the rule is that the leading coefficient on the squared term is a positive 1.  Ours is a -2, so we have to factor it out. $-2( x^{2} -4x)=5$.  Complete the square by taking half the linear term (the linear term is 4 now), square it and add it to both sides.  Half of 4 is 2, and 2 squared is 4, so we will add it in.  HOWEVER, when you add it into the parenthesis on the left, you still have that -2 hanging out front, so what you really have added in is -2(4) which is -8.  Here's what we have now so far: $-2( x^{2} -4x+4)=5-8$ which simplifies to $-2( x^{2} -4x+4)=-3$.  The next step is to create the perfect square binomial we formed during this process on the left, which gives us $-2(x-2) ^{2}=-3$.  Move the -3 back over to the other side by addition and you get $-2(x-2) ^{2} +3=y$.  This tells us that our vertex is (2, 3)