PARABOLA
I)
Parabola 1 : (1, -1), (3, 13), (-2, -7)
Parabola 2 : -x2+x+7
(1) a + b + c = -1
(2) 9a + 3b + c = 13
(3) 4a - 2b + c = -7
(1)-(2) : (4) 8a + 2b = 14
(2)-(3) : (5) 5a +5b = 20
(4)-(5) : a = 1 , b = 3 , c = -5
So, parabola 1 : y = x2+3x-5
Intersection :
y=y
x2+3x-5= -x2+x+7
2x2+2x+-12 =0
2(x+3)(x-2) =0
x= -3 or 2 , y= -5 or 5
So, point of intersection (x,y) = (-3,-5), (2,5)
Line :
y-5=2(x+1)
2x-y+1=0
Parabola 1
Parabola 2
x-intercept
x= 1.2 or -4.2
x int, = (1.2,0) and
(-4.2,0)
y-intercept
Axis of symmetry
Vertex
f(0)=5
y int. = (0,5)
x= -b/2a= -1.5
(1.5,f(1.5))=(1.5,1.75)
x= -2.2 or -3.2
x int, = (-2.2,0) and
(3.2,0)
f(0)=7
y int. = (0,7)
x= -b/2a=0.5
(0.5,f(0.5))=(0.5,7.75)
PARABOLA
II).
Parabola 1 : (-1,1), (2,-14), (3,-23)
Parabola 2 : y = 3x2+16x+22
(1) a - b + c = 1
(2) 4a + 2b + c = -14
(3) 9a + 3b + c = -23
(1)-(2) : (4) 3a + 3b =-15
(2)-(3) : (5) 5a + b = -9
(4)-(5) : a = -1 , b = -4 , c = -2
So, parabola 1 : y = -x2-4x-2
Intersection :
y=y
-x2-4x-2 = 3x2+16x+22
4x2+20x+24 =0
x2+5x+6
(x+2)(x+3) =0
x= -2 or -3 , y= 2 or 1
So, point of intersection (x,y) = (-2,2), (-3,1)
Line :
𝑦−2
𝑥−(−2)
=
2−1
−2−(−3)
=
1
1
y-2 = x +2
x-y+4=0
Parabola 1
Parabola 2
x-intercept
x= or
x int, = (,0) and
(,0)
y-intercept
Axis of symmetry
Vertex
f(0)=-2
y int. = (0,-2)
x= -b/2a=-2
(2,f(2))=(2,-14)
x= or
x int, = (,0) and
(,0)
f(0)=22
y int. = (0,22)
x= -b/2a= -2.7
(-2.7,f(-2.7))=(2.7,87)
INTERSECTION OF TWO PARABOLAS
Parabola 1 : (0,2), (1,9), (3,35)
Parabola 2 : -x2+2x+2
(1) c = 2
(2) a + b + c = 9
(3) 9a + 3b + c = 35
(1)-(2) : (4) a + b = 7
(1)-(3) : (5) 9a + 3b = 33
(4)-(5) : a =2 , b = 5 , c = 2
So, parabola 1 : y = 2x2+5x+2
Intersection :
y=y
2x2+5x+2= -x2+2x+2
3x2+3x =0
3x(x+1) =0
x= -1 or 0 , y= -1 or 2
So, point of intersection (x,y) = (-1,-1), (0,2)
Line :
𝑦−2
𝑥−0
=
−1−2
−1−0
=3
y-2=3x
3x+y-2=0
Parabola 1
Parabola 2
x-intercept
x= or
x int, = (,0) and
(,0)
y-intercept
Axis of symmetry
Vertex
f(0)=2
y int. = (0,2)
x= -b/2a=-1.25
(-1.25,f(-1.25))=1.25,1)
x= or
x int, = (,0) and
(,0)
f(0)=2
y int. = (0,2)
x= -b/2a=1
(1,f(1))=(1,3)
INTERSECTIONS OF TWO PARABOLAS
Parabola 1 : (1,7), (-2,19), (-1,13)
Parabola 2 : -2x2+9x
(1) a + b + c = 7
(2) 4a - 2b + c = 19
(3) a - b + c = 13
(1)-(2) : (4) 3a - 3b = 12
(2)-(3) : (5) 3a + b = 7
(4)-(5) : a = 1 , b = -3 , c = 9
So, parabola 1 : y = x2-3x+9
Intersection :
y=y
x2-3x+9= -2x2+9x
3x2-12x+9 =0
3(x-1)(x-3) =0
x= 1 or 3 , y= 7 or 9
So, point of intersection (x,y) = (1,7), (3,9)
Line :
𝑦−7
𝑥−1
=
9−7
3−1
2
= =1
2
y-7=x-1
x-y+6=0
Parabola 1
Parabola 2
No x-intercepts
x= 0 or 4.5
x int, = (0,0) and
(4.5,0)
f(0)= 9
y int. = (0,9)
x= -b/2a=1.5
f(0)=0
y int. = (0,0)
x= -b/2a= 2.25
x-intercept
y-intercept
Axis of
symmetry
Vertex
(1.5,f(1.5))=(1.5,6.75) (2.25,f(2.25))=(2.25,10.3)
MATH REMEDIAL
Parabola 1 : (-1,-8), (1,-2), (2,4)
Parabola 2 : -2x2+12
(1) a - b + c = -8
(2) a + b + c = -2
(3) 4a + 2b + c = 4
(1)-(2) : (4) 2b = 6
(2)-(3) : (5) 3a + b = 6
(4)-(5) : a = 1 , b = 3 , c = -6
So, parabola 1 : y = x2+3x-6
Intersection :
y=y
-2x2+12= x2+3x-6
3x2+3x-18 =0
3(x+3)(x-2) =0
x= -3 or 2 , y=-6 or 4
So, point of intersection (x,y) = (-3,-6), (2,4)
Line :
𝑦−4
𝑥−2
=
−6−4
−3−2
=2
y-4=2x-4
2x-y=0
Parabola 1
Parabola 2
x-intercept
x= -4.4 or 1.4
x int, = (-4.4,0) and
(1.4,0)
y-intercept
Axis of symmetry
Vertex
f(0)= -6
y int. = (0,-6)
x= -b/2a=-1.5
(-1.5,f(-1.5))=(-1,5,8.25)
x= -2.5 or 2.5
x int, = (-2.5,0) and
(2.5,0)
f(0)=12
y int. = (0,12)
x= -b/2a=0
(0,f(0))=(0,12)
MATH REMEDIAL
Parabola 1 : (2,23), (1,11), (-2,-1)
Parabola 2 : -3x2+x+3
(1) 4a + 2b + c = 23
(2) a + b + c = 11
(3) 4a - 2b + c = -1
(1)-(2) : (4) 3a + b = 12
(1)-(3) : (5) 4b = 24
(4)-(5) : a =2 , b =6 , c = 3
So, parabola 1 : y = 2x2+6x+3
Intersection :
y=y
2x2+6x+3 = -3x2+x+3
5x2+5x =0
5x(x+1) =0
x= 0 or -1 , y= 3 or -1
So, point of intersection (x,y) = (0,3), (-1,-1)
Equation of the Line :
𝑦−3
𝑥−0
=
−1−3
−1−0
=4
y-3=4x
4x-y+3=0
Parabola 1
Parabola 2
x-intercept
x=-0.6 or -2.4
x int, = (-0.6,0) and
(-2.4,0)
y-intercept
Axis of symmetry
Vertex
f(0)=3
y int. = (0,3)
x= -b/2a=-1.5
(-1.5,f(-1.5))=(-1.5,1.5)
x=-0.8 or 1.2
x int, = (-0.8,0) and
(1.2,0)
f(0)=3
y int. = (0,3)
x= -b/2a=0.2
(0.2,f(0.2))=(0.2,3.1)
MATH
Parabola 1 has three points (-1,3), (0,9), (1,17)
Parabola 2 : -2x2-11x-15
(1) a - b + c = 3
(2) c = 9
(3) a + b + c = 17
(1)-(2) : (4) a - b = -6
(2)-(3) : (5) a + b = 8
(4)-(5) : a =1 , b = 7 , c = 9
So, parabola 1 : y = x2+7x+9
Intersection :
y=y
x2+7x+9= -2x2-11x-15
3x2+18x+24 =0
3(x+2)(x+4) =0
x= -2 ,-4 , y= -1,-3
point of intersection (x,y) = (-2,-1), (-4,-3)
Line :
𝑦−(−3)
𝑥−(−4)
=
−3−(−1)
−4−(−2)
=
−2
−2
= −1
y+3=x+4
x-y+1=0
Parabola 1
Parabola 2
x-intercept
x= -5.3 or -1.7
x int, = (-5.3,0) and
(-1.7,0)
y-intercept
Axis of symmetry
Vertex
f(0)=9
y int. = (0,9)
x= -b/2a=3.5
(3.5,f(3.5))=(3.5,3.25)
x= -3 or -2.5
x int, = (-3,0) and
(-2.5,0)
f(0)=
y int. = (0,)
x= -b/2a=-2.8
(-2.8,f(-2.8))=(2.8,0.1)
MATH
Parabola 1 : (2,12), (0,2), (-2,8)
Parabola 2 : -x2+7x+2
(1) 4a + 2b + c = 12
(2) c = 2
(3) a - b + c = 8
(1)-(2) : (4) 4a + 2b = 10
(2)-(3) : (5) a - b = 6
(4)-(5) : a =2 , b = 1 , c =2
So, parabola 1 : y = 2x2+x+2
Intersection :
y=y
2x2+x+2= -x2+7x+2
3x2-6x =0
3x(x-2) =0
x= 0 or 2 , y= 2 or 12
point of intersection (x,y) = (0,2), (2,12)
Line :
𝑦−2
𝑥−0
=
12−2
2−0
=5
y-2=5x
5x-y+2=0
Parabola 1
Parabola 2
No x intercepts
x=-0.3 or -7.3
x int, = (-0.3,0) and
(-7.3,0)
f(0)=
y int. = (0,2)
x= -b/2a=0.3
(=0.3,f(0.3))=(0.3,1.9)
f(0)=2
y int. = (0,2)
x= -b/2a=3.5
(3.5,f(3.5))=(3.5,14.3)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH
Parabola 1 : (1,9), (2,14), (3,21)
Parabola 2 : -2x2-x+24
(1) a + b + c = 9
(2) 4a + 2b + c = 14
(3) 9a + 3b + c = 21
(1)-(2) : (4) 3a + b = 5
(2)-(3) : (5) 5a + b = 7
(4)-(5) : a = 1 , b = 2 , c = 6
So, parabola 1 : y = x2+2x+6
Intersection :
y=y
x2+2x+6= -2x2-x+24
3x2+3x-18 =0
3(x+3)(x-2) =0
x= -3or 2 , y=9 or 14
So, point of intersection (x,y) = (-3,9), (2,14)
Line :
𝑦−14
𝑥−2
=
9−14
−3−2
=
−5
−5
=1
y-14=x-2
x-y+12=0
Parabola 1
Parabola 2
NO x intercepts
x= -3.7 or 3.2
x int, = (,0) and
(,0)
f(0)=6
y int. = (0,6)
xs= -b/2a=-1
(xs,f(xs))=(=1,5)
f(0)=24
y int. = (0,24)
xs= -b/2a=0.3
(xs,f(xs))=(0.3,24.1)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH
Parabola 1 : (1,8), (-1,-4), (3,28)
Parabola 2 : -x2+2x+1
(1) a + b + c = 8
(2) a - b + c = -4
(3) 9a + 3b + c = 28
(1)-(2) : (4) 2b = 12
(2)-(3) : (5) 8a + 4b = 32
(4)-(5) : a = 1 , b = 6, c = 1
So, parabola 1 : y = x2+6x+1
Intersection :
y=y
x2+6x+1= -x2+2x+1
2x2+4x =0
2x(x+2) =0
x= 0 or -2 , y=1 or -7
So, point of intersection (x,y) = (0,1), (-2,-7)
Line :
𝑦−1
𝑥−0
=
−7−1
−2−0
=4
y-1=4x
4x-y+1=0
Parabola 1
Parabola 2
x-intercept
x= -5.8 or 0.1
x int, = (,0) and
(,0)
y-intercept
Axis of symmetry
Vertex
f(0)=
y int. = (0,)
x= -b/2a=3
(xs,f(xs))=(3,8)
x= or
x int, = (,0) and
(,0)
f(0)=
y int. = (0,)
x= -b/2a=
(,f())=(,)
MATH REMEDIAL
Parabola 1 : (0,5), (1,4), (2,13)
Parabola 2 : -2x2+7x-1
(1) c = 5
(2) a + b + c = 4
(3) 4a + 2b + c = 13
(1)-(2) : (4) a + b = -1
(1)-(3) : (5) 4a + 2b = 8
(4)-(5) : a = 1 , b = -2 , c = 5
So, parabola 1 : y = x2-2x+5
Intersection :
y=y
x2-2x+5= -2x2+7x-1
3x2-9x+6=0
3(x-1)(x-2) =0
x= 1, 2 , y= 4, 5
point of intersection (x,y) = (1,4), (2,5)
Line :
𝑦−4
𝑥−1
=
5−4
2−1
=1
y-4=x-1
x-y+3=0
Parabola 1
Parabola 2
No x intercepts
x=0.1 or 4/
x int, = (,0) and
(,0)
f(0)=5
y int. = (0,5)
xs= -b/2a=1
(xs,f(xs))=(1,4)
f(0)=-1
y int. = (0,-1)
xs= -b/2a=1.75
(xs,f(xs))=(1.75,5.1)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH REMEDIAL
Parabola 1 : (-1,-3), (1,3), (2,-3)
Parabola 2 : 2x2+3x-2
(1) a - b + c = -3
(2) a + b + c = 3
(3) 4a + 2b + c = -3
(1)-(2) : (4) 2b = 6
(2)-(3) : (5) 3a + b = -6
(4)-(5) : a = -3 , b = 3 , c = 3
So, parabola 1 : y = -3x2+3x+3
Intersection :
y=y
-3x2+3x+3= 2x2+3x-2
5x2-5 =0
5(x+1)(x-1) =0
x= -1 or 1 , y= -3 or 3
So, point of intersection (x,y) = (-1,-3), (1,3)
Line :
𝑦−3
𝑥−1
=
−3−3
−1−1
=3
y-3=3x--3
3x-y=0
Parabola 1
Parabola 2
x-intercept
x= -0.6 or 1.6
x int, = (-0.6,0) and
(1.6,0)
y-intercept
Axis of symmetry
Vertex
f(0)=3
y int. = (0,3)
xs= -b/2a=0.5
(xs,f(xs))=(0.5,3.75)
x=-2 or 0.5
x int, = (-2,0) and
(0.5,0)
f(0)=-2
y int. = (0,-2)
x= -b/2a=-0,75
(xs,f(xs))=(-0/.5,-3.1)
MATH REMEDIAL
Parabola 1 : (0,0), (1,4), (2,14)
Parabola 2 : x2-x+4
(1) c = 0
(2) a + b + c = 4
(3) 4a + 2b + c = 14
(1)-(2) : (4) a + b = 4
(2)-(3) : (5) 3a + b = 10
(4)-(5) : a =3 , b =1 , c =0
So, parabola 1 : y = 3x2+x
Intersection :
y=y
3x2+x+= x2-x+4
2x2+2x-4 =0
2(x+2)(x-1) =0
x= -2 or 1 , y= -2 or 4
So, point of intersection (x,y) = (-2,-2), (1,4)
Line :
y-4= 2(x-1)
2x-y+2=0
Parabola 1
Parabola 2
x-intercept
No x interepts
x= 0 or -0.3
x int, = (0,0) and
(-0.3,0)
y-intercept
Axis of symmetry
Vertex
f(0)=0
y int. = (0,0)
x= -b/2a=-0.2
(xs,f(xs))=(-0.2,-0.08)
f(0)=4
y int. = (0,4)
x= -b/2a=0.5
(,f())=(0.5,3.75)
MATH REMEDIAL
Parabola 1 : (-1,5), (0,3), (1,5)
Parabola 2 : x2-x+5
(1) a - b + c = 5
(2) c = 3
(3) a + b + c = 5
(1)-(2) : (4) a - b = 2
(2)-(3) : (5) a + b = 2
(4)-(5) : a = 2 , b = 0 , c = 3
So, parabola 1 : y = 2x2+3
Intersection :
y=y
2x2+3= x2-x+5
x2+x-2 =0
(x+2)(x-1) =0
x= -2 or 1 , y= -1 or 5
So, point of intersection (x,y) = (-2,-1), (1,5)
Line :
y-5=2(x-1)
2x-y+3=0
Parabola 1
Parabola 2
No x intercepts
x= or
No x intercepts
x-intercept
y-intercept
Axis of symmetry
Vertex
f(0)=3
y int. = (0,3)
x= -b/2a=0
(,f())=(0,3)
f(0)=5
y int. = (0,5)
xs= -b/2a=0.5
(xs,f(xs))=(,0.5,
4.75)
MATH REMEDIAL
Parabola 1 : (-1,8), (0,6), (1,6)
Parabola 2 : -x2-3x+10
(1) a - b + c = 8
(2) c = 6
(3) a + b + c = 6
(1)-(2) : (4) a - b = 2
(2)-(3) : (5) a + b = 0
(4)-(5) : a = 1 , b = -1 c = 6
So, parabola 1 : y = x2-x+6
Intersection
y=y
-x2-3x+10 = x2-x+6
2x2+2x-4 = 0
2(x+2)(x-1) =0
x= -2 or 1 , y= 0 or 6
So, point of intersection (x,y) = (-2,0), (1,6)
Line :
y-6=2(x-1)
2x-y+4=0
Parabola 1
Parabola 2
No x intercepts
x= -5 or 2
x int, = (-5,0) and
(2,0)
f(0)=6
y int. = (0,6)
Xs= -b/2a=0.5
(xs,f(xs))=(0.5,6.76)
f(0)= 10
y int. = (0,10)
xs= -b/2a=-1.5
(xs,f(xs))=(-1.5,12.25)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH REMEDIAL
Parabola 1 : (-2,2) (2,6), (1,2)
Parabola 2 : -x2+5x
(1) 4a - 2b + c = 2
(2) 4a + 2b + c = 6
(3) a + b + c = 2
(1)-(2) : (4) 4b = 4
(2)-(3) : (5) 3a + b = 4
(4)-(5) : a = 1 , b = 1 , c = 0
So, parabola 1 : y = x2+x
Intersection :
y=y
-x2+5x = x2+x
2x2-4x =0
2x(x-2) =0
x= 0 or 2 , y= 2 or 6
So, point of intersection (x,y) = (0,2), (2,8)
Line :
y-6=3(x-2)
3x-y=0
Parabola 1
Parabola 2
x-intercept
x= -1 or 0
x int, = (0,0) and
(-1,0)
y-intercept
Axis of symmetry
Vertex
f(0)=0
y int. = (0,0)
xs= -b/2a=-0.5
(xs,f(xs))=(-0.5,)
x= 0 or 5
x int, = (0,0) and
(5,0)
f(0)=0
y int. = (0,0)
xs= -b/2a=2.5
(xs,f(xs))=(2.5,6.25)
MATH REMEDIAL
Parabola 1 : (-1,-4), (0,-1), (1,4)
Parabola 2 : -x2+4x+1
(1) a - b + c = -4
(2) c = -1
(3) a + b + c = 4
(1)-(2) : (4) a - b = -3
(2)-(3) : (5) a + b = 5
(4)-(5) : a = 1 , b = 4 , c = -1
So, parabola 1 : y = x2+4x-1
Intersection :
y=y
-x2+4x+1= x2+4x-1
2x2-2=0
2(x+1)(x-1) =0
x= -1 or 1 , y= -4 or 4
So, point of intersection (x,y) = (-1,-4), (1,4)
Line :
y-4=4(x-1)
4x-y=0
Parabola 1
Parabola 2
x-intercept
x= -4.2 or 0.2
x int, = (-4.2,0) and
(0.2,0)
y-intercept
Axis of symmetry
Vertex
f(0)=-1
y int. = (0,-1)
xs= -b/2a=-2
(xs,f(xs))=(-2,-5)
x=-0.2 or 4.2
x int, = (-0.2,0) and
(4.2,0)
f(0)=1
y int. = (0,1)
xs= -b/2a=2
(xs,f(xs))=(2,5)
MATH REMEDIAL
Parabola 1 : (-1,-2), (0,1), (1,6)
Parabola 2 : -2x2+4x+4
(1) a - b + c = -2
(2) c = 1
(3) a + b + c = 6
(1)-(2) : (4) a - b = -3
(2)-(3) : (5) a + b = 5
(4)-(5) : a = 1 , b = 4 , c = 1
So, parabola 1 : y = x2+4x+1
Intersection :
y=y
-2x2+4x+4 = x2+4x+1
3x2-3x =0
3(x+1)(x-1) =0
x= -1 or 1 , y= -2 or 6
So, point of intersection (x,y) = (-1,-2), (1,6)
Line :
y-6=4(x-1)
4x-y+2=0
Parabola 1
Parabola 2
x-intercept
x= -3.7 or -0.3
x int, = (-3.7,0) and
(-0.3,0)
y-intercept
Axis of symmetry
Vertex
f(0)=1
y int. = (0,1)
xs= -b/2a=-2
(xs,f(xs))=(-2,-3)
x= -0.7 or 2.7
x int, = (-0,7,0) and
(2.7,0)
f(0)=4
y int. = (0,4)
xs= -b/2a=1
( xs,f ( xs))=(1,6)
MATH REMEDIAL
Parabola 1 : (-1,-1), (0,2), (1,7)
Parabola 2 : -3x2+4x+6
(1) a - b + c = -1
(2) c = 2
(3) a + b + c = 7
(1)-(2) : (4) a - b = -3
(2)-(3) : (5) a + b = 5
(4)-(5) : a = 1 , b = 4 , c =2
So, parabola 1 : y = x2+4x+2
Intersection :
y=y
-3x2+4x+6= x2+4x+2
4x2-4x =0
4(x+1)(x-1) =0
x= -1 or 1 , y= -1 or 7
So, point of intersection (x,y) = (-1,-1), (1,7)
Line :
y-7=3(x-1)
4x-y+3=0
Parabola 1
Parabola 2
x-intercept
x= -3.4 or -0.6
x int, = (-3.4,0) and
(-0.6,0)
y-intercept
Axis of symmetry
Vertex
f(0)=2
y int. = (0,2)
xs= -b/2a=-2
(xs,f(xs))=(-2,-2)
x= -0.9 or 2.2
x int, = (-0.9,0) and
(2.2,0)
f(0)=6
y int. = (0,6)
xs= -b/2a=0.7
( xs,f( xs))=(0.7,7.3)
MATH REMEDIAL
Parabola 1 : (-1,-3), (0,2), (1,5)
Parabola 2 : 2x2+4x-1
(1) a - b + c = -3
(2) c = 2
(3) a + b + c = 5
(1)-(2) : (4) a - b =-5
(2)-(3) : (5) a + b = 3
(4)-(5) : a = -1 , b = 4, c = 2
So, parabola 1 : y = -x2+4x+2
Intersection :
y=y
2x2+4x-1= -x2+4x+2
3x2-3x =0
3(x+1)(x-1) =0
x= -1 or 1 , y= -3 or 5
So, point of intersection (x,y) = (-1,-3), (1,5)
Line :
y-5=4(x-1)
4x-y+1=0
Parabola 1
Parabola 2
x-intercept
x= 0.4 or 4.4
x int, = (0.4,0) and
(4.4,0)
y-intercept
Axis of symmetry
Vertex
f(0)=2
y int. = (0,2)
xs= -b/2a=2
(xs,f(xs))=(2,6)
x= -2.2 or 0.2
x int, = (-2.2,0) and
(0.2,0)
f(0)=-1
y int. = (0,-1)
xs= -b/2a=-1
(xs, f( xs))=(-1,-3)
MATH REMEDIAL
Parabola 1 : (0,1), (1,3), (2,7)
Parabola 2 : -2x2+7x+1
(1) c = 1
(2) a + b + c =3
(3) 4a + 2b + c = 7
(1)-(2) : (4) a + b = 2
(2)-(3) : (5) 3a + b = 4
(4)-(5) : a =1 , b = 1 , c = 1
So, parabola 1 : y = x2+x+1
Intersection :
y=y
-2x2+7x+1= x2+x+1
3x2-6x =0
3x(x-2) =0
x= 0 or 2 , y= 1 or 7
So, point of intersection (x,y) = (0,1), (2,7)
Line :
y-7=3(x-2)
3x-y+1=0
Parabola 1
Parabola 2
No x intercept
x= 0.1 or 3.6
x int, = (0.1,0) and
(3.6,0)
f(0)=1
y int. = (0,1)
xs= -b/2a=-0.5
(xs,f( xs))=(-0.5,0.75)
f(0)=1
y int. = (0,1)
xs= -b/2a=1.75
(xs ,f(xs))=(1.75,7.1)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH REMEDIAL
Parabola 1 : (0,2), (1,11), (2,8)
Parabola 2 : -2x2+7x+2
(1) c = 2
(2) a + b + c = 6
(3) 4a + 2b + c = 8
(1)-(2) : (4) a + b = 4
(2)-(3) : (5) 3a + b = 2
(4)-(5) : a = -1 , b = 5 , c = 2
So, parabola 1 : y = -x2+5x+2
Intersection :
y=y
-2x2+7x+2= -x2+5x+2
x2-2x =0
x(x-2) =0
x= 0 or 2 , y= 2 or 8
So, point of intersection (x,y) = (0,2), (2,8)
Line :
y-8=3(x-2)
3x-y+2=0
Parabola 1
Parabola 2
x-intercept
x= -0.4 or 5.4
x int, = (-0.4,0) and
(5.4,0)
y-intercept
Axis of symmetry
Vertex
f(0)=2
y int. = (0,2)
xs= -b/2a=2.5
(xs,f xs))=(2.5,8.3)
x= 0.3 or 3.8
x int, = (0.3,0) and
(3.8,0)
f(0)=2
y int. = (0,2)
xs= -b/2a=1.75
(xs,f( xs))=(1.75,8.1)
MATH REMEDIAL
Parabola 1 : (0,3), (1,4), (2,9)
Parabola 2 : 3x2-3x+3
(1) c = 3
(2) a + b + c = 4
(3) 4a + 2b + c = 9
(1)-(2) : (4) a + b = 1
(2)-(3) : (5) 3a + b = 5
(4)-(5) : a = 2 , b = -1 , c = 3
So, parabola 1 : y = 2x2-1x+3
Intersection :
y=y
3x2-3x+3= 2x2-x+3
x2-2x =0
x(x-2) =0
x= 0 or 2 , y= 3 or 9
So, point of intersection (x,y) = (0,3), (2,9)
Line :
y-9=3(x-2)
3x-y+3=0
Parabola 1
Parabola 2
No x intercept
x= or
No x intercept
f(0)=3
y int. = (0,3)
xs= -b/2a=0.25
( xs,f(xs))=(0.25,2.9)
f(0)=3
y int. = (0,3)
xs= -b/2a=0.5
(xs,f( xs))=(0.5,2.25)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH REMEDIAL
Parabola 1 : (0,4), (1,6), (2,10)
Parabola 2 : -3x2+9x+4
(1) c = 4
(2) a + b + c = 6
(3) 4a + 2b + c = 10
(1)-(2) : (4) a + b = 2
(2)-(3) : (5) 3a + b = 4
(4)-(5) : a = 1 , b = 1 , c = 4
So, parabola 1 : y = x2+x+4
Intersection :
y=y
-3x2+9x+4= x2+x+4
4x2-8x =0
4x(x-2) =0
x= 0 or 2 , y= 4 or 10
So, point of intersection (x,y) = (0,4), (2,10)
Line :
y-10=3(x-2)
3x-y+4=0
Parabola 1
Parabola 2
No x intercepts
x= -0.4 or 0.4
x int, = (-0.4,0) and
(0.4,0)
f(0)=4
y int. = (0,4)
xs= -b/2a=-0.5
(xs,f(xs))=(-0.5,3.75)
f(0)=4
y int. = (0,4)
xs= -b/2a=1.5
(xs,f(xs))=(1.5,10,75)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH REMEDIAL
Parabola 1 : (-1,2), (0,-1), (1,0)
Parabola 2 : x2-1
(1) a - b + c = 2
(2) c = -1
(3) a + b + c = 0
(1)-(2) : (4) a - b = 3
(2)-(3) : (5) a + b = 1
(4)-(5) : a = 2, b = -1, c = -1
So, parabola 1 : y = 2x2-x-1
Intersection :
y=y
x2-1= 2x2-x-1
x2-x =0
x(x-1) =0
x= 0 or 1 , y= -1 or So, point of intersection (x,y) = (0,-1), (1,0)
Line :
y-0=x-1
x-y-1=0
Parabola 1
Parabola 2
x-intercept
x= -0.5 or 1
x int, = (-0.5,0) and
(1,0)
y-intercept
Axis of symmetry
Vertex
f(0)=-1
y int. = (0,-1)
xs= -b/2a=0.25
(xs,f(xs))=(0.25,1.1)
x= -1 or 1
x int, = (-1,0) and
(1,0)
f(0)=-1
y int. = (0,-1)
xs= -b/2a=0
(xs,f(xs))=(0,1)
MATH REMEDIAL
Parabola 1 : (-1,3), (0,0), (1,1)
Parabola 2 : -x2+2x
(1) a - b + c = 3
(2) c = 0
(3) a + b + c = 1
(1)-(2) : (4) a - b = 3
(2)-(3) : (5) a + b = 1
(4)-(5) : a = 2 , b = -1 , c = 0
So, parabola 1 : y =2x2-x
Intersection :
y=y
-x2+2x=2x2-x
3x2-3x =0
3x(x-1) =0
x= 0 or 1 , y= 1 or 2
So, point of intersection (x,y) = (0,0), (1,1)
Line :
y-1=x-1
x-y=0
Parabola 1
Parabola 2
x-intercept
x= 0 or 0.5
x int, = (9,0) and
(0.5,0)
y-intercept
Axis of symmetry
Vertex
f(0)=0
y int. = (0,0)
xs= -b/2a=0.25
(xs,f(xs))=(0.25,0.125)
x= 0 or 2
x int, = (0,0) and
(2,0)
f(0)=0
y int. = (0,0)
xs= -b/2a=1
(xs,f(xs))=(1,1)
MATH REMEDIAL
Parabola 1 : (-1,6), (0, 1), (1,2)
Parabola 2 : -3x2+4x+1
(1) a - b + c = 6
(2) c = 1
(3) a + b + c = 2
(1)-(2) : (4) a - b = 5
(2)-(3) : (5) a + b = 1
(4)-(5) : a = 3 , b = -2 , c = 1
So, parabola 1 : y = 3x2-2x+1
Intersection :
y=y
-3x2+4x+1= 3x2-2x+1
6x2-6x =0
6x(x-1) =0
x= 0 or 1 , y= 1 or 2
So, point of intersection (x,y) = (0,1), (1,2)
Line :
y-2=x-1
x-y+1=0
Parabola 1
Parabola 2
No x intercepts
x= 0.2 or 1.5
x int, = (0.2,0) and
(1.5,0)
f(0)=1
y int. = (0,1)
xs= -b/2a=0.3
(xs,f(xs))=(0.3,0.7)
f(0)=1
y int. = (0,1)
xs= -b/2a=0.7
(xs,f(xs))=(0.7,2.3)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH REMEDIAL
Parabola 1 : (-1,3), (0,2), (1,3)
Parabola 2 : 2x2-x+2
(1) a - b + c = 3
(2) c = 2
(3) a + b + c = 3
(1)-(2) : (4) a - b = 1
(2)-(3) : (5) a + b = 1
(4)-(5) : a = 1 , b = 0, c = 2
So, parabola 1 : y =x2+2
Intersection :
y=y
2x2-x+2= x2+2
x2-x =0
x(x-1) =0
x= 0 or 1 , y= 2 or 3
So, point of intersection (x,y) = (0,2), (1,3)
Line :
y-3=x-1
x-y+2=0
Parabola 1
Parabola 2
No x intercept
x= or
x int, = (,0) and
(,0)
f(0)=3
y int. = (0,3)
x= -b/2a=0
(xs,f(xs))=(0,2)
f(0)=
y int. = (0,)
x= -b/2a=0.25
(xs,f(xs))=(0.25,1.9)
x-intercept
y-intercept
Axis of symmetry
Vertex
MATH REMEDIAL
Parabola 1 : (-1,6), (0,3), (1,4)
Parabola 2 : 3x2-2x+3
(1) a - b + c = 6
(2) c = 3
(3) a + b + c = 4
(1)-(2) : (4) a - b = 3
(2)-(3) : (5) a + b = 1
(4)-(5) : a = 2 , b = -1 , c = 3
So, parabola 1 : y = 2x2-x+3
Intersection :
y=y
3x2-2x+3= 2x2-x+3
x2-x =0
x(x-1) =0
x= 0 or 1 , y= 3 or 4
So, point of intersection (x,y) = (0,3), (1,4)
Line :
y-4=x-1
x-y+3=0
Parabola 1
Parabola 2
x-intercept
No x intercept
No x intercept
y-intercept
Axis of symmetry
Vertex
f(0)=3
y int. = (0,3)
xs= -b/2a=0.3
(xs,f(xs))=(0.3,-2.9)
f(0)=3
y int. = (0,3)
xs= -b/2a= 0.3
(xs,f(xs))=(0.3,.7)
MATH REMEDIAL
Parabola 1 : (-1,5), (0,4), (1,5)
Parabola 2 : -x2+2x+4
(1) a - b + c = 5
(2) c =4
(3) a + b + c = 5
(1)-(2) : (4) a - b = 1
(2)-(3) : (5) a + b = 1
(4)-(5) : a = 1 , b = 0 , c = 4
So, parabola 1 : y = x2+4
Intersection :
y=y
-x2+2x+4= x2+4
2x2-2x =0
x(x-1) =0
x= 0 or 1 , y= 4 or 5
So, point of intersection (x,y) = (0,4), (1,5)
Line :
y-5=x-1
x-y+4=0
Parabola 1
Parabola 2
NO x intercept
x= 1.2 or 3,2
x int, = (1.2,0) and
(3.2,0)
f(0)=4
y int. = (0,4)
xs= -b/2a=0
(xs,f(xs))=(0,4)
f(0)=4
y int. = (0,4)
xs= -b/2a=1
(xs,f(xs))=(1,5)
x-intercept
y-intercept
Axis of symmetry
Vertex