graphing parabolas. pattern probes 1. which of the following relationships describes the pattern:...
TRANSCRIPT
GRAPHING PARABOLAS
Pattern Probes
1. Which of the following relationships describes the pattern: 1,-1,-3,-5…?
A. tn = -2n
B. tn = -2n +3
C. tn = 2n – 3
D. tn = 2n
Pattern Probes
2. What is the 62th term of the pattern 1,9,17,25,…?
A. 489B. 496C. 503D. 207
A linear pattern is given by tn = t1 +(n-1)d
Quadratics Patterns
Last time we learned that:
A quadratic pattern is defined by tn = an2 + bn +c
The CD = 2a for a quadratic
A quadratic pattern has its CD on D2.
A linear pattern has a common difference on D1.
Note: Another phrase for linear pattern is an arithmetic sequence.
Let’s Graph (Put on your graphing shoes and graph in blue?)
Quadratics, as we know, do not increase by the same amount for every increase in x. (How do we know this?)Therefore, their graphs are not straight lines, but curves. In fact, if you’ve ever thrown a ball, watched a frog jump, or taken Math 12, you’ve seen a quadratic graph!
Consider the function f(x) = x2 - the simplest quadratic (Why?)
The pattern would be
Its table of values would be
y = x2
x y
-3 9
-2 4
-1 1
0 0
1 1
2 4
3 9
Graphing Cont
x y
-3 9
-2 4
-1 1
0 0
1 1
2 4
3 9
We can plot these points to get the PARABOLA
Vertex
Once we have the vertex, the rest of the points follow the pattern:Over 1, up 1Over 2, up 4Over 3, up 9
Axis of Symmetry: x = 0
Domain: xε RRange: y ε [0,∞) or y ≥ 0
Transformations of y= x2
That parabola (which represents the simplest quadratic function), can be
Stretched vertically
Slid horizontally
Reflected
Slid vertically These are called transformations: Ref, VS, HT, VT,
These show up (undone) in the transformational form of the equation:
21HTxVTy
VSThe form y = ax2 + bx + c is
called the GENERAL FORM
TransformationsEx. Graph the function
2)1(32
1 xy
What are the transformations?
Reflection: No (the left side is not multiplied by -1)
Vertical Stretch, VS? Yes VS = 2
Vertical Translational, VT? Yes VT = -3
Horizontal Translational, HT? Yes HT = 1
So we can use these to get a MAPPING RULE:
(x,y) (x+1, 2y-3)
We now take the old table of values and do what the mapping rule says.
TransformationsEx. Graph the function
2)1(32
1 xy
So we can use these to get a MAPPING RULE:
(x,y) (x+1, 2y-3)
We now take the old table of values and do what the mapping rule says.
y = x2
x y
-3 9
-2 4
-1 1
0 0
1 1
2 4
3 9
½(y+3) = (x-1)2
x y
-2 15
-1 5
0 -1
1 -3
2 -1
3 5
4 15
TransformationsEx. Graph the function
2)1(32
1 xy
½(y+3) = (x-1)2
x y
-2 15
-1 5
0 -1
1 -3
2 -1
3 5
4 15
y = x2
x y
-3 9
-2 4
-1 1
0 0
1 1
2 4
3 9
New Vertex (1,-3) that is (HT, VT)
TransformationsEx. Graph the function
2)1(32
1 xy
½(y+3) = (x-1)2
x y
-2 15
-1 5
0 -1
1 -3
2 -1
3 5
4 15
New Vertex (1,-3) that is (HT, VT)
Once we find the new vertex, the rest of the points follow the patternOver 1, up 1 x VS Over 2, up 4 x VSOver 3, up 9 x VS
Domain: x ε RRange: y ε [-3,∞)Axis of symmetry: x = 1
Graph the following. State the axis of symmetry, range, domain and the y-intercept of each. A)
B)
C)
2)2(32
1 xy
2)1(2 xy
2)6()4( xy
A) 2)2(32
1 xy
Ref: NoVS: 2VT: 3HT: 2
(x,y)(x+2,2y+3)
½(y-3) = (x-2)2
x y
-1 21
0 11
1 5
2 3
3 5
4 11
5 21
Domain: x ε RRange: y ε [3,∞)Axis of symmetry: x = 2Y-int (0,11)
B)
Ref: YesVS: 1/2VT: 0HT: -1
(x,y)(x-1,-0.5y)
-2y = (x+1)2
x y
-4 -4.5
-3 -2
-2 -0.5
-1 0
0 -0.5
1 -2
2 -4.5
Domain: x ε RRange: y ε (-∞,0]Axis of symmetry: x = -1Y-int (0,-0.5)
2)1(2 xy
C)
Ref: YesVS: 1VT: 4HT: -6
(x,y)(x-6,-y+4)
-(y-4) = (x+6)2
x y
-9 -5
-8 0
-7 3
-6 4
-5 3
-4 0
-3 -5
Domain: x ε RRange: y ε (∞,4]Axis of symmetry: x = -6Y –int (0,-32)
2)6()4( xy
Going backwards
Find the equation of this graph:
Ans: 2(y+1)=(x-1)2
Going Backwards
a. What quadratic function has a range of y ε(-∞,6], a VS of 4 and an axis of symmetry of x = -5?
b. What quadratic function has a vertex of (2,-12) and passes through the point (1,-9)?a) 2)5()6(
4
1 xy b) 2)2()12(
3
1 xy
Solving for x using transformational form
We can use transformational form to solve for x- values given certain y-values.We can use transformational form to solve for x- values given certain y-values.
Example: The parabola
2)3()6(2
1 xy
passes through the point (x,2). What are the values of x?
Solving for x using transformational form
Find the x-intercepts of the following quadratic functions:Find the x-intercepts of the following quadratic functions:
2)8()5(4 xy 2)1()2(2
1 xy
x
x
x
x
x
208
820
)8(20
)8(20
)8()50(4
2
2
2
11
)1(1
)1(1
)1()20(2
1
2
2
2
x
x
x
x
Changing Forms
General Form: y = ax2 +bx + cTransformational Form:
Standard Form:
2)()(1
hxkya
khxay 2)(
What are you good for?
General Form: f(x)=y = ax2 +bx + c2)()(
1hxky
aTransformational Form:
Good for:
Getting the mapping ruleGetting the mapping rule
Graphing the function’s parabolaGraphing the function’s parabola
Draw backs: Functions are usually written in the f(x) notation.
Getting the equation of the graphGetting the equation of the graph
Good for: Finding the y-intercept
(0,c)
Finding the y-intercept
(0,c)Finding the x-intercept(s)
(soon)
Finding the x-intercept(s)
(soon)
Finding the vertex (soon)
Finding the vertex (soon)
khxay 2)(
Standard Form:
Good for: Nothing
Going To General Form
Going To General Form
2)3( x
)3)(3( xx
2x x3 x3 9
96)3( 22 xxx
Going to General Form
Ex. Put the following into general form:
2)4()3(2
1 xy
)4)(4()3(2
1 xxy
168)3(2
1 2 xxy
)168(2)3(2
12 2
xxy
321623 2 xxy
29162 2 xxy