F4 MATHS[2](聽日TEST啦,唔該各位幫幫我,有20點!!)

3(a)
sub ( -1,-11)
-11=-4(-1-2)^2+k
k=-11+36
k=25
(b)
y= -4(x-2)² + 25
sub y=0
-4(x-2)² + 25=0
(x-2)^2=25/4
x=5/2+2 or x=-5/2+2
x=9/2 or -1/2
4
(a)
The graph of y=a(x-h)² + k passes through A( 1, -3/2 ) and cuts the x-axis at B(2,0) and C(4,0)
sub A( 1, -3/2 ) ,B(2,0) and C(4,0)
-3/2=a(1-h)^2+k
0=a(2-h)^2+k
0=a(4-h)^2+k
so
a(2-h)^2=a(4-h)^2
(2-h)^2=(4-h)^2
2-h=-(4-h)
2-h=h-4
2h=6
h=3
we get
-3/2=4a+k
0=a+k
so
a=-k
-3/2=-4k+k
-3k=-3/2
k=1/2
a=-1/2
(b)
y=a(x-h)² + k
y=-1/2(x-3)² + 1/2
when x=3, y gets the maximum value 1/2
(c)
sub x=0
y-intercept
=-1/2(-3)^2+1/2
=-4
5 (a + 2)² = a² + 4 ???? correct or wrong?why?
(a+2)^2
=a^2+4a+4
since 4a may not be equal to 0
so the identity is wrong

3(a)

sub ( -1,-11)

-11=-4(-1-2)^2+k

k=-11+36

k=25

(b)

y= -4(x-2)² + 25

sub y=0

-4(x-2)² + 25=0

(x-2)^2=25/4

x=5/2+2 or x=-5/2+2

x=9/2 or -1/2

4

(a)

The graph of y=a(x-h)² + k passes through A( 1, -3/2 ) and cuts the x-axis at B(2,0) and C(4,0)

sub A( 1, -3/2 ) ,B(2,0) and C(4,0)

-3/2=a(1-h)^2+k

0=a(2-h)^2+k

0=a(4-h)^2+k

so

a(2-h)^2=a(4-h)^2

(2-h)^2=(4-h)^2

2-h=-(4-h)

2-h=h-4

2h=6

h=3

we get

-3/2=4a+k

0=a+k

so

a=-k

-3/2=-4k+k

-3k=-3/2

k=1/2

a=-1/2

(b)

y=a(x-h)² + k

y=-1/2(x-3)² + 1/2

when x=3, y gets the maximum value 1/2

(c)

sub x=0

y-intercept

=-1/2(-3)^2+1/2

=-4

5 (a + 2)² = a² + 4 ???? correct or wrong?why?

(a+2)^2

=a^2+4a+4

since 4a may not be equal to 0

so the identity is wrong