what information can we visually determine from this quadratic graph? vertex vertical stretch factor...
TRANSCRIPT
What information can we visually determine from this Quadratic Graph?
–2 –1 1 2 3 4 5 6 7 8 9 10
–4
–3
–2
–1
1
2
3
4
x
y
Vertex
Vertical Stretch Factor
(4, -3)
Over 1, up 1
Over 2, up 4
Normal Pattern
so VS=1
2)()(1
pxqya
2)4()3( xySo the function is
What information can we visually determine from this Quadratic Graph?
Vertex
Vertical Stretch Factor
(-2, 1)
Over 1, up ?
Over 2, up ?
Over 3, up 3 ( )
so VS=
2)()(1
pxqya
2)2()1(3 xySo the function is
–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4
5
10
x
y
9x3
1
3
1
What about this Quadratic Graph?
Vertex
Vertical Stretch Factor
(1, -5)
Over 1, down 2
Over 2, down 8
so VS= -2
2)()(1
pxqya
2)1()5(2
1 xySo the function is
)1x2(
–2 –1 1 2 3 4
–20–19–18–17–16–15–14–13–12–11–10–9–8–7–6–5–4–3–2–1
12
xy
)4x2(
Now it’s your turn!Determine the Functions that model each of these curves:
–10 –5 5 10
–10
–8
–6
–4
–2
2
4
6
8
10
x
y
Curve B
Curve A 27x y
2) 2 ( ) 3 ( 2 x y
Determine the Functions that model each of these curves:
Curve B 2) 8 ( ) 3 ( x y
–10 –5 5 10
–10
–8
–6
–4
–2
2
4
6
8
10
x
y
Curve A 2) 5 ( ) 2 ( 2 x y
2 Sad Parabolas
Why?
a < 1
Determine the Functions that model each of these curves:
Curve B
Curve A 2) 4 ( 8 x y
2) 5 ( ) 9 (
5
1 x y
–10 –5 5 10
–10
–8
–6
–4
–2
2
4
6
8
10
x
y
2 Happy Parabolas
Why?
a > 1
Remember, quadratic functions represent real-life situations such as:
Real Life Connection
…the height of a kicked soccer ball
y = (-9.8)t2 + 20t
…the motion of falling objects pulled by gravity:
height = -16t2 + 100
Graphs of Quadratic Functions
Quadratic Functions graph into a shape called a
Parabola
If the vertical stretch, “a”, is negative, the
parabola opens downward
If the vertical stretch, “a”, is positive, the
parabola opens upward
Vertex
Maximum Point
Minimum Point
2)5( xy 2)4( xy
Determining Functions of Parabolas ALGEBRAICALLY!
1. Find the equation of the parabola having a vertex of (3, 5) passing through the point (1, 2).
)5,3(),( qp)2,1(),( yx
2)31()52(1
a
SUBSTITUTE
2)2()3(1
a
4)3(1
a
3
41
a
The FUNCTION for this parabola is:
2)3()5(3
4 xy
Determining Functions of Parabolas ALGEBRAICALLY!
2. Find the equation of the parabola having a vertex of (-1, 4) with a y-intercept of –3.
)4,1(),( qp)3,0(),( yx
2)10()43(1
a
SUBSTITUTE
2)1()7(1
a
1)7(1
a
7
11
a
The FUNCTION for this parabola is:
2)1()4(7
1 xy
Determining Functions of Parabolas ALGEBRAICALLY!
3. A quadratic function has x-intercepts (-1, 0) and (5, 0). Find the formula for the function if it has a maximum value of 6.
)6,2(),( qp)0,5(),( yx
2)25()60(1
a
SUBSTITUTE
2)3()6(1
a
9)6(1
a
2
31
a
The FUNCTION for this parabola is:
2)2()6(2
3 xy
The axis of symmetry is located halfway between the two intercepts so x=2 is the axis of symmetry.
Determining Functions of Parabolas ALGEBRAICALLY!
4. A bridge is shaped as a parabolic arch. If the maximum height is 20 m and the bridge spans 60 m, find the function that models this situation. )20,30(),( qp
)0,0(),( yx
2)300()200(1
a
900)20(1
a
451
a
The FUNCTION for this model is:
2)30()20(45 xy10 20 30 40 50 60
10
20
x
y
20 m
60 m
Homework: Page 38 #39. 42, 43, 44