中學的數學問題!!

when y = 0,
2x^2 + 4x = 0
=> x = 0 or x = -2
X-intercepts are 0,0 and -2,0

when x = 0
y = 0
Y-intercept is 0,0

d(y)/d(x) = 0 is the turning point
=> 4x + 4 = 0
=> x = -1
when x = -1, y = -2
vertex is -1,-2

If you use the maths method,

let y=2x^2+4x=0

vertex = -b/2a = -4/[2(2)] = -1
so vertex x-axis = -1
put x=-1 into y= 2x^2+4x
y = 2(-1)^2+4(-1)
= -2
so the curve vertex is (-1,-2)

y-intercept is put x=0 into y= 2x^2+4x
y = 2(0)^2+4(0) = 0

x-intercepts is put y=0 into y= 2x^2+4x
2x^2+4x = 0
x^2+2x = 0
x(x+2) = 0
x=0 or x=-2