Soal :
1)
y = x2 – 25
2)
y = -x2 + 7x – 12
3)
y = -x2 – 8x + 9
4)
y = -2x2 – 7x + 4
5)
y = -x2 +6x
JAWABAN
1. y = x2 – 25
Titik potong sumbu –x ( y = 0)
y = x2 – 25
0 = x2 – 25
( x – 5 ) ( x + 5)
x = 5 x= -5
koordinat titik potong sumbu –x (5,0) dan (-5,0)
Titik potong sumbu –y( x = 0 )
y = x2 – 25
y = 0 – 25
y = -25
koordinat titik potong sumbu –y ( 0, -25)
Koordinat titik puncak (xp,yp)
xp= -b/2a
xp= -0/2 . 1
xp= -0/2
xp= 0
yp= -b2-4ac / 4a
yp= -02 - 4.1.-25 /4.1
yp= -0 + 100 /4
yp= -25
koordinat titik puncak (xp,yp) = ( 0, -25 )
2. y = –x2 + 7x – 12
Titik potong sumbu –x ( y = 0)
y = –x2 + 7x – 12
0 = -x2 + 7x -12
0 = x2 - 7x +12
( x – 4) ( x – 3 )
x = 4 x = 3
koordinat titik potong sumbu –x ( 4,0) dan ( 3,0 )
Titik potong sumbu –y( x = 0 )
y = -x2 + 7x -12
y = -02 + 0 -12
y = -12
koordinat titik potong sumbu –y ( 0,-12 )
Koordinat titik puncak (xp,yp)
xp= -b/2a
xp= -7/2.-1
xp= 7/2
yp= -b2-4ac / 4a
yp= -72-4.-1. -12/ 4.-1
yp= -49-48 / -4
yp= 97 / 4
koordinat titik puncak (xp,yp) ( 7/2, 97/4 )
3. y = –x2 – 8x + 9
Titik potong sumbu –x ( y = 0)
y = –x2 – 8x + 9
0 =-x2 - 8x +9
x2 + 8x -9 = 0
( x + 9 ) ( x – 1 )
x = - 9 x = 1
koordinat titik potong sumbu –x ( - 9,0 ) dan ( 1,0)
Titik potong sumbu –y( x = 0 )
Y =-x2 - 8x +9
0 =-02 - 0 +9
Y = 9
koordinat titik potong sumbu –y ( 0,9)
Koordinat titik puncak (xp,yp)
xp= -b/2a
xp= -(-8) /2.-1
xp= -4
yp= -b2-4ac / 4a
yp= -(-8)2-4.-1. 9/ 4.-1
yp= -64+36 / -4
yp= -100 / -4
yp= 25
koordinat titik puncak (xp,yp) (-4, 25 )
4. y= –2x2 – 7x + 4
Titik potong sumbu –x ( y = 0)
y = -2x2-7x + 4
0 = -2x2-7x + 4
2x2+7x - 4 = 0
1/2 (2x + 8) ( 2 x – 1 )
( x + 4 ) ( 2 x – 1 )
x = -4 2x = 1
x = 1/2
koordinat titik potong sumbu –x ( -4, 0) dan ( 1/2, 0 )
Titik potong sumbu –y( x = 0 )
y = -2x2-7x + 4
y = 0-0+ 4
y = 4
koordinat titik potong sumbu –y ( 0,4)
xp = -(b/2a)
= -(-7/2.-2)
=-(-7/-4)
= - 1,75
yp = -(b2 – 4ac/4a)
=- (72 – 4.-2.4/4.-2)
= -(49 + 32/-8)
= -(-10,125)
= 10,125
koordinat titik puncak (xp,yp) (-1,75; 10,125)
5. y = –x2 + 6x
Titik potong sumbu –x ( y = 0)
y = -x2 + 6x
0 = -x2 + 6x
x2 - 6x =0
x ( x - 6 )
x = 0 x = 6
koordinat titik potong sumbu –x ( 0, 0) dan ( 6, 0 )
Titik potong sumbu –y( x = 0 )
y = -x2 + 6x
y = -02 + 0
y = 0
koordinat titik potong sumbu –y ( 0,0)
Langkah 3
Koordinat titik puncak (xp,yp)
xp= -b/2a
xp= -6/2.-1
xp= 3
yp= -b2-4ac / 4a
yp= -62-4.-1.0 / 4.-1
yp= -36-0 / -4
yp= 9
koordinat titik puncak (xp,yp) ( 3,9)
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