F.4 Maths 2題

1)
y - interxept = -1 (sub x=0)
x-intercept (sub y=0)
2x^2 - 4x - 1=0

x = {-(-4) ±√[(-4)^2 - 4(2)(-1)]} / 2(2)
= (2 ± √6)/2

x-intercepts are (2 +√6)/2 and (2 -√6)/2

2)
y- intercept = -5
x- intercept:
-2x^2 + 8x -5 =0
x = {-(8) ±√[(8)^2 - 4(-2)(-5)]} / 2(-2)
= (4 ± √6)/2

x-intercepts are (4 +√6)/2 and (4 -√6)/2

2010-09-29 00:48:24 補充：
1) Coordinates of y-intercept is (0,-1)
Coordinates of x-intercepts are ( (2 +√6)/2, 0),( (2 -√6)/2, 0)

2) Coordinates of y-intercept is (0,-5)
Coordinates of x-intercepts are ( (4 +√6)/2, 0),( (4 -√6)/2, 0)

其實中四既數唔suppose你禁機
所以最好寫番exact value instead of rounded off value, e.g. 寫(2 +√6)/2唔寫2.25

2010-09-30 23:34:49 補充：
If the question asked for y-intercept, you may answer -1; for coordinates of y-intercept, you should answer (0,-1).

You are supposed to calculate the roots using the quadratic formula, thus the result should have the square root sign in it e.g.(2 ± √6)/2, instead of the approximate value (e.g. 2.225)

1.) y=2x^2-4x-1
x-intercepts =(x,0)
y=0
0=2x^2-4x-1
x=[4±√(16+8)]/4
x=1±√(24/16)
x=1±√(3/2)
x=2.225 or -0.225
x-intercepts=(-0.225,0) and (2.225,0)
y-intercepts =(0,y)
y=2x^2-4x-1
y=-1
y-intercepts =(0,-1)

2.) y=-2x^2+8x-5
x-intercepts (x,0)
0=-2x^2+8x-5
0=[-8±√(64-40)]/-4
0=2±√(3/2)
x=3.225 or -0.775
x-intercepts is (3.225,0) and (-0.775,0)
y-intercepts (0,y)
y=-2x^2+8x-5
y=-5
y-intercepts is (0,-5)