F. 5 maths
2015-04-12 6:13 am
1.Write the equation of the quadraticfunction, first in vertex form and then in standard form.a. vertex(-4, 8) and passing through (2, -4)b. vertex(-6, -5) and passing through (-3, 4) 2. A quadratic function has zero at 1 and –5and passes through the poin (2, 10), write the equation in vertex form.
回答 (2)
2015-04-12 7:37 am
✔ 最佳答案
1.(a)
Let y = a(x + 4)² + 8 be the vertex form of the requiredequation.
(2, -4) lies on the graph of the equation :
-4 = a(2 + 4)² + 8
36a = -12
a = -1/3
y = (-1/3)(x + 4)² + 8
y = (-1/3)(x + 4)² + 8
3y = -(x + 4)² + 24
3y = -x² - 8x - 16 + 24
Standard form : y = -(1/3)x² - (8/3)x + (8/3)
(b)
Let y = a(x + 6)² - 5 be the vertex form of the requiredequation.
(-3, 4) lies on the graph of the equation :
4 = a(-3 + 6)² - 5
9a = 9
a = 1
y = (x + 6)² - 5
y = (x + 6)² - 5
y = x² + 12x + 36 - 5
Standard form : y = x² + 12x + 31
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2.
Let y = ax² + bx + c
When x = 1, y = 0 :
0 = a(1)² + b(1) + c
a + b + c = 0 ...... [1]
When x = -5, y = 0 :
0 = a(-5)² + b(-5) + c
25a - 5b + c = 0 ...... [2]
When x = 2, y = 10 :
10 = a(2)² + b(2) + c
4a + 2b + c = 10 ...... [3]
[2] - [1] :
24a - 6b = 0
4a - b = 0 ...... [4]
[3] - [1] :
3a + b = 10 ..... [5]
[4] + [5] :
7a = 10
a = 10/7
Put into [4] :
4(10/7) - b = 0
b = 40/7
Put the values of a and b into [1] :
(10/7) + (40/7) + c = 0
c = -50/7
y = (10/7)x² + (40/7)x - (50/7)
y = (10/7)(x² + 4x) - (50/7)
y = (10/7)(x² + 4x + 4) - 4(10/7) - (50/7)
vertex form : y = (10/7)(x + 2)² - (90/7)
2015-04-12 8:13 am
1a
y=a(x+4)^2 +8
put (x,y)=(2,-4)
-4=a(2+4)^2 +8
a=-1/3
y=(-1/3)(x+4)^2 +8
y=(-1/3)(x^2 +8x+16)+8
y=(-x^2 -8x-16+24)/3
y=(-x^2 -8x+8)/3
1b
y=a(x+6)^2 +5
put (x,y)=(4,-5)
y=a(x+6)^2 -5
a=1
y=(x+6)^2 -5
y=x^2 +12x+36-5
y=x^2 +12x+31
2
y=ax^2 -bx+c
put (x,y)=(0,1)
a+b+c=0 i
put (x,y)=(0,-5)
25a-5b+c=0 ii
put (x,y)=(2,10)
4a+2b+c=10 iii
a=10/7
b=40/7
c=-50/7
y=(10/7)x^2 +(40/7)x-(50/7)
h=-b/(2a)
h=-(40/7)/(2*10/7)
h=-2
k=(4ac-b^2)/4a
k=[4(10/7)(-50/7)-(40/7)^2]/(4*10/7)
k=-90/7
y=(10/7)(x+2)^2 -90/7
y=a(x+4)^2 +8
put (x,y)=(2,-4)
-4=a(2+4)^2 +8
a=-1/3
y=(-1/3)(x+4)^2 +8
y=(-1/3)(x^2 +8x+16)+8
y=(-x^2 -8x-16+24)/3
y=(-x^2 -8x+8)/3
1b
y=a(x+6)^2 +5
put (x,y)=(4,-5)
y=a(x+6)^2 -5
a=1
y=(x+6)^2 -5
y=x^2 +12x+36-5
y=x^2 +12x+31
2
y=ax^2 -bx+c
put (x,y)=(0,1)
a+b+c=0 i
put (x,y)=(0,-5)
25a-5b+c=0 ii
put (x,y)=(2,10)
4a+2b+c=10 iii
a=10/7
b=40/7
c=-50/7
y=(10/7)x^2 +(40/7)x-(50/7)
h=-b/(2a)
h=-(40/7)/(2*10/7)
h=-2
k=(4ac-b^2)/4a
k=[4(10/7)(-50/7)-(40/7)^2]/(4*10/7)
k=-90/7
y=(10/7)(x+2)^2 -90/7
收錄日期: 2021-04-15 19:09:32
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https://hk.answers.yahoo.com/question/index?qid=20150411000051KK00134