1.If the graph y=x^2+4 is translated 3 units to
the right and 2 units upward, what will be the
equation of y=x^2+4?
2.f(x)=6k-(k+2)x such that f(5)=-5.k=?(show steps)
3.What is the axis of symmetry of the parabola
y=-2x^2+12x+5?
4.Solve the equation(x+3)(x+a)=x+3.
5.If k<0, the equation x^2-2x-k+2=0 has
A.a repeated root.
B.no real root.
C.two distinct rational roots.
D.two distinct irrational roots.(Show steps and explain)
Maths.
2007-12-22 10:50 pm
回答 (2)
2007-12-22 11:05 pm
✔ 最佳答案
As follows:圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths057.jpg?t=1198307100
2007-12-22 15:09:17 補充：
Sorry for some typing errors in Q3, y = -2x^2 + 12x + 5 = - 2 ( x^2 - 6x ) + 5= -2 ( x^2 - 6x + 9 ) + 5 + 2 ( 9 )= - 2 ( x - 3 )^2 + 23The axis of symmetry: x = 3
參考: My Maths Knowledge
2007-12-22 11:26 pm
1.y=(x-4)^2+6
2.Put x=5 into f(x),
f(5)=6k-(k+2)5
-5k=6k-(k+2)5
k=5/3
so,f(5)= -25/3
3.completing square,
y=-2x^2+12x+5
=-2(x^2-6x)+5
=-2(x^2-6x+9)+5+2(9)
==-2(x-3)^2+23
so,x=3 is the axis of symmetry.
4.(x+3)(x+a)=x+3
(x+3)(x+a)-(x+3)=0
(x+3)(x+a-1)=0
x=-3 or x=1-a
5.answer is B.
consider the discriminant,
discriminant = (-2)^2-4(-k+2)
= 4+4k-8
=4k-4
=4(k-1)
as k<0, 4(k-1)<0,
there is no real root for the equation.
2007-12-22 15:29:23 補充：
amendment for Q.22.Put x=5 into f(x), f(5)=6k-(k 2)5-5=6k-(k 2)5k=5
2.Put x=5 into f(x),
f(5)=6k-(k+2)5
-5k=6k-(k+2)5
k=5/3
so,f(5)= -25/3
3.completing square,
y=-2x^2+12x+5
=-2(x^2-6x)+5
=-2(x^2-6x+9)+5+2(9)
==-2(x-3)^2+23
so,x=3 is the axis of symmetry.
4.(x+3)(x+a)=x+3
(x+3)(x+a)-(x+3)=0
(x+3)(x+a-1)=0
x=-3 or x=1-a
5.answer is B.
consider the discriminant,
discriminant = (-2)^2-4(-k+2)
= 4+4k-8
=4k-4
=4(k-1)
as k<0, 4(k-1)<0,
there is no real root for the equation.
2007-12-22 15:29:23 補充：
amendment for Q.22.Put x=5 into f(x), f(5)=6k-(k 2)5-5=6k-(k 2)5k=5
收錄日期: 2021-04-26 13:18:22
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