中四附加數

As follows~~~


圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/si.jpg?t=1191931091

a)because the parabola touches the x-axis.
so, △= 0
6^2-4(2)(k)= 0
36-8k = 0
-8k = -36
k = 9/2

b) dy/dx=4x+6=0
x=-3/2
d^2y/dx^2=4>0(maximum)
maximum point(-3/2,0)

c)whenx=0
y-intercept=9/2

d) y=2x^2+6x+k
y=x+k
2x^2+6x+k=x+k
2x^2 +5x = 0
x(2x + 5) =0
x=0 or x= -5/2

a) The parabola touches the x-axis, ie only one root
so △= (6)^2 - 4(2)(k) = 0
36 -8k = 0
k = 9/2
= 4.5

b) By completing square,
y=2x^2+6x+k
= 2(x^2 + 3x + (3/2)^2 - (3/2) ^2 ) +k
= 2(x+ 3/2)^2 +k -9/2

so y = k -9/2 = 9/2 - 9/2 = 0
where x = -3/2
vertex = ( -3/2 , 0)

c) The equation of a parabola is now: y=2x^2+6x+9/2
Put x=0 into the equation,
y= 9/2

so, y-intercept of the parabola = 9/2

d) y=2x^2+6x+k and y=x+k

so, x+ k = 2x^2+6x+k
2x^2 +5x = 0
x(2x + 5) =0
x = 0 or x= -5/2

when x=0, y=0+9/2 = 9/2

when x= -5/2, y= -5/2 + 9/2 = 2

a : y=2x^2+6x+k
The parabola touches the x-axis.
so, △> 0
6^2-4(4)(k)> 0
36-16k > 0
-16k > -36
k > 36/16