Quadratic Equations show/explain work?

Question

I need help finding the y-intercepts, and the vertexs.

1) find the vertex and the y-intercept.
y=2x²+4x+2

2) find the vertex and the y-intercept.
y=-3x²+12x-9

How do you find it? Show/work explain please.
3)
For the 1) I got 2(x²+2x+1) is y=2(x+1)²+1 for the vertex. Which is (1,-1) but its wrong why?

Answer 1

1) find the y-intercept.
let x= 0
y=2x²+4x+2
=> y = 2

&
find the vertex
y=2x²+4x+2
=> dy/dx = 4x + 4
=> 4x + 4 = 0
=> x = 1
&
y=2(1)²+4(1)+2
=> y = 8
=> vertex = ( 1,8 )

Answer 2

1.
Factorise the eq'n 2x^2 + 4x + 2 = 2(x^2 + 2x + 1)
Reading the eq'n , The y-intercept is '+2; the terminal free standing number; if 'x' is 0(zero) then the only value 'y' can have is '+2' - the intercept.
The vertex is by taking the term in 'x' (+2x) halving the coefficient and change the sign so it becomes '-1'. This is the coordinate for 'x' at the vertex. To find the coordinate for 'y' take '-1' and substitute it in to the full eq'n ;-
2(-1)^2 +4(-1) +2
2 - 4 + 2 = 0
So the vertex is at (-1,0).
Using Differential Calculus to find the vertex.
y = 2x^2 +4x + 2
dy/dx = 4x + 4
At the vertex the gradient is zero so
dy/dx = 0
therefore
4x + 4 = 0
4x = -4
x = -1
Again substituting '-1' for x into the full eq'n we have y = 0
So the vertex is at (-1,0)
NB When you factored,the second stage is incorrect. y is not equal to 2(x+1)^2 + 1 the 'plus one' is the incorrect bit.

2.
y = 3x^2 +12x -9
y = 3(x^2 + 4x - 3)
Vertex 'x' = -2
therefore vertex 'y' = -21

Or y = 3x^2 + 12x -9
dy/dx = 6x + 12
6x + 12 = 0
6x = -12
x = -2
and by substitution for 'x'
y = 3(-2)^2 +12(-2) -9
y = 12 -24 -9
y = -21
Vertex is at (-2, -21)

Similarly, the y-intercept is '-9' when 'x' is 0 (zero).