Question

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

(2, 3)
(2, 19)
(2, 27)

Answer 1

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)

Answer 2

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)