A parabola has the vertex at (1,3) and it cuts the y-axis at (0,5).
a)Find the equation of the parabola
b)If the line y=mx touches the parabila, find the value(s) of the constant m.
c)If the parabola is shifted horizontally so that the y-axis becomes its line of symmetry, what will be the new equation of the parabola?
F.4 AM
2007-10-04 3:45 am
回答 (2)
2007-10-04 4:05 am
✔ 最佳答案
a) Let the equation of the parabola be a(x-1)²+3=0
a(0-1)²+3=5
a=2
so, the equation of the parabola is 0=2(x-1)²+3
0=2x²-4x+5
b) y=mx
y=2x²-4x+5
put y=mx into y=2x²-4x+5
mx=2x²-4x+5
0=2x²-4x-mx+5----------(*)
delta of * =0 (it touches the parabola)
(-4-m)²-4x2x5=0
m²+8m-24=0
m= -4加減開方10
c) parabola has to move to left 1 unit
y=2x²-4x+5
new parabola : y=2(x+1)²-4(x+1)+5
y=2x²+4x+2-4x-4+5
y=2x²+3
2007-10-14 19:32:45 補充：
sorry
i've made a mistake
m= -4加減開方10
m應該係-4加減2乘開方10
2007-10-04 4:12 am
a)y=ax^2+bX+c
c=5
a(x^2+bx/2a)^2+c-b^2/4a (((completing square)))
咁佢vertex係1,3.....姐係話b/2a = -1
c-b^2/4a = 3
b/2a = -1
b^2 / 4a = 2
b = -2a
(-2a)^2/4a = 2
a = 2
b = -4
y = 2x^2 - 4x +5
b) 2x^2 - 4x + 5 = mx
2x^2 - x(4+m)+5 = 0
D = 16+8m + m^2 -40 = 0
m= -8 +/- root(64+4*24) over 2
m=-4 +/- 2root(10)
c) 本身個條既completing square 係 2(x-1)^2 + 3
咁宜家個y-axis becomes its line of symmetry
姐係變左 y=2(x-0)^2 + 3
化返開姐係變左 y = 2x^2 +3
c=5
a(x^2+bx/2a)^2+c-b^2/4a (((completing square)))
咁佢vertex係1,3.....姐係話b/2a = -1
c-b^2/4a = 3
b/2a = -1
b^2 / 4a = 2
b = -2a
(-2a)^2/4a = 2
a = 2
b = -4
y = 2x^2 - 4x +5
b) 2x^2 - 4x + 5 = mx
2x^2 - x(4+m)+5 = 0
D = 16+8m + m^2 -40 = 0
m= -8 +/- root(64+4*24) over 2
m=-4 +/- 2root(10)
c) 本身個條既completing square 係 2(x-1)^2 + 3
咁宜家個y-axis becomes its line of symmetry
姐係變左 y=2(x-0)^2 + 3
化返開姐係變左 y = 2x^2 +3
參考: me
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