find the vertex?

Question

Find the vertex of the parabola which is the graph of the quadratic function y=-2x^2+4x+6

Answer 1

I factored out -2 first to make this easier to see.

y = -2(x² - 2x - 3)

Take the number in front of x (-2), divide it by 2 (-2/2 = -1), square it (1), and both add it and subtract it:

y = -2(x² - 2x + 1 - 1 - 3)

Now the first three terms x² - 2x + 1 are a perfect square (x - 1)², and you're left with -4 on the right.

y = -2((x - 1)² - 4)

Now to get it in the right form to find the vertex, go ahead and distribute the -2:

y = -2(x - 1)² + 8

This is in the correct form, y = a(x - h)² + k, so the vertex is:

(1, 8).

I'll double check by differentiating. Don't worry if you don't understand this, it's just a different way to do it.

y' = -4x + 4 = 0, so -4x = -4, so x = 1, check.
y(1) = -2(1)² + 4(1) + 6 = -2 + 4 + 6 = 8, check.

So I'm sure it's right now.

Answer 2

The axis of symmetry is x = -b/2a = -4/2(-2)= 1
So y = -2+ 4 +6 =10
The vertex which lies on the axis of symmetry is (1,10).

Answer 3

take the first derivative and equal it to zero:
y'=-4x+4=0
x=1
substitute this value in the first equation
y=-2(1)^2+4(1)+6=8

The vertex is (1,8)

Answer 4

x = -1
y = 4

this coordinate is the lowest point of the parabola.

Differentiate the function and set = to zero and solve for x.

Then use this x to solve for y in the original equation

Answer 5

no idea, sorry

Answer 6

answer= (-1,4)