Graphing Quadratic Functions:Parabolas
2. What is the 62th term of the pattern:1, 9, 17, 25, …?
A. 489B. 496C. 503D. 207
1. Which of the following relationships describes the pattern:1, −1, −3, −5 …?
A. tn = −2nB. tn = −2n + 3C. tn = 2n – 3D. tn = 2n
Pattern Probes
1. Which of the following relationships describes the pattern:1, −1, −3, −5 …?
A. tn = −2nB. tn = −2n + 3C. tn = 2n – 3D. tn = 2n
Pattern Probes2. What is the 62th term of
the pattern:1, 9, 17, 25, …?
A. 489B. 496C. 503D. 207
Quadratics Patterns
Last time we learned that:
A linear pattern has a common difference on level 1.A linear pattern is given by tn = t1 + (n − 1)d
A quadratic pattern has its common difference on level 2.The CD = 2a for a quadraticThe quadratic pattern is defined by tn = an2 + bn + cThe general form of the quadratic function is defined by
f(x) = ax2 + bx + c
Note: Another phrase for linear pattern is an arithmetic sequence.
Let’s Graph
Quadratics, as we know, do not increase by the same amount for every increase in x (or n). (How do we know this?)Therefore, their graphs are not straight lines,but curves, called parabolas. In fact, if you’veever thrown a ball, or watched a frog jumpyou’ve seen a quadratic graph!
Consider the functionf(x) = x2
It’s the simplest quadratic (Why?)
a = 1b = 0c = 0
y = x2
x y–3 9
–2 4
–1 1
0 0
1 1
2 4
3 9
Its table of values would be:
Let’s Graph
Quadratics, as we know, do not increase by the same amount for every increase in x (or n). (How do we know this?)Therefore, their graphs are not straight lines,but curves, called parabolas. In fact, if you’veever thrown a ball, or watched a frog jumpyou’ve seen a quadratic graph!
Consider the functionf(x) = x2
It’s the simplest quadratic (Why?)
The pattern would be
term #, (n) 1 2 3 4 5value, (tn) 1 4 9 16 25
Its table of values would be:
a = 1b = 0c = 0
y = x2
x y–3 9
–2 4
–1 1
0 0
1 1
2 4
3 9
Let’s Graph
Once we have the vertex, the rest of the points follow the pattern:over 1 (from the vertex), up 1 (from the vertex)
over 2 (from the vertex), up 4 (from the vertex)
over 3 (from the vertex), up 9 (from the vertex)
over 7 (from the vertex), up 49 (from the vertex)
y = x2
x y–3 9–2 4–1 10 01 1 2 43 9
Properties of the graph y = x2
axis of symmetry:the line x = 0
vertex:the point (0, 0)
domain:{x R}
range:{y ≥ 0, y R}
Transformations of y = x2
This simplest parabola can be:
We give these transformations short forms: VS, HT, and VT,
These show up (undone) in the transformational form of the equation:
2HTVTVS1
xy
TransformationsEx. Graph the function:(FYI, its general formis y = 2x2 – 4x – 1)
2)1(321
xy
What are the transformations that have been applied to the base graph of y = x2?
Vertical StretchVS = 2 (positive, so no reflection)
Vertical TranslationalVT = –3
Horizontal TranslationalHT = 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.
old x and y values, from y = x2
new x and y values
TransformationsEx. Graph the function:(FYI, its general formis y = 2x2 – 4x – 1)
2)1(321
xy
Apply the mapping rule:(x, y)→ (x + 1, 2y – 3)
y = x2
x y–3 9–2 4–1 10 01 1 2 43 9
“old” x and y values(from the base parabola)
½(y + 3) = (x – 1)2
x y–2 15–1 50 –11 –32 –13 54 15
New x and y values for this transformed parabola
TransformationsEx. Graph the function: 2)1(3
21
xy
½(y + 3) = (x – 1)2
x y–2 15–1 50 –11 –32 –13 54 15
Transformationson y = x2:VS = 2VT = –3HT = 1
TransformationsEx. Graph the function: 2)1(3
21
xy
Short-cut:Once we have the vertex,
the rest of the points follow the pattern:
over 1, up 1 × 2over 2, up 4 × 2over 3, up 9 × 2
.
.
.over #, up # × VS
(again, always from the vertex)
TransformationsEx. Graph the function: 2)1(3
21
xy
vertex (1, –3)(that is (HT, VT))
axis of symmetry x = 1(that is the line x = HT)
domain {x R}(unchanged)
range {y ≥ –3, y R}(that is {y ≥ VT, y R })
Graphing Parabolas: PracticeGraph the following. State the vertex, axis of symmetry, range,
domain and the y-intercept of each.
A) (or y = 2x2 – 8x + 11, in general form)
B) (or y = – ½x2 – x – ½ , in general form)
C) (or y = –x2 – 12x – 32, in general form)
2)2(321
xy
2)1(2 xy
2)6()4( xy
2)2(321
xy (or y = 2x2 – 8x + 11 in general form)
½(y – 3) = (x – 2)2
x y
-1 21
0 11
1 5
2 3
3 5
4 11
5 21
VS = 2HT = 2VT = 3
(x, y) → (x + 2, 2y + 3)
domain: {x R}range: {y ≥ 3, y R}axis of symmetry: x = 2y-intercept (0,11)
A)
(or y = -½x2 - x - ½ , in general form)
VS = –½HT = –1VT = 0
(x, y) → (x – 1, – ½y)
domain: {x R}range: {y ≤ 0, y R}axis of symmetry: x = –1y-intercept (0, – ½)
2)1(2 xy
-2y = (x + 1)2
x y
-4 -4.5
-3 -2
-2 -0.5
-1 0
0 -0.5
1 -2
2 -4.5
B)
(or y = –x2 – 12x – 32 in general form)
½(y – 3) = (x – 2)2
x y
-1 21
0 11
1 5
2 3
3 5
4 11
5 21
VS = –1 HT = –6VT = 4
(x, y) → (x – 6, – y + 4)
domain: {x R}range: {y ≤ 4, y R}axis of symmetry: x = –6y-intercept (0, –32)
2)6()4( xyC)
Going backwards: Finding the Equation
Answer:2(y + 2)=(x – 1)2
Find the equation of this graph:
Give the general form of this equation:
23
21 2 xxy
More on this to come….
a. What quadratic function has a range of {y ≥ 6, y R}, 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, –15)?
a) 2)5()6(41
xy b) 2)2()12(31
xy
Going backwards: Finding the Equation
Answers
Example:The parabola
2)3()6(21
xy
passes through the points (x, 2). What are the values of x?
We can use transformational form to solve for the x-values given a certain y-value.(Note – there will (almost) always be TWO such x-values!)
1,53234
34
3821
36221
)3()6(21
2
2
2
2
xxxx
x
x
x
xy
Solving for x given yusing transformational form
2)8()5(4 xy 2)1()2(21
xy
x
x
x
x
x
528
208
820
820
85042
2
11
11
)1(1
12021
2
2
x
x
x
x
Solving for x given yusing transformational form
Find the x-intercepts (roots) of the following quadratic functions:
2)1()2(21
xy
422
)1(22
)1(221
1)0(221
2
2
2
yy
y
y
y
Solving for y given xusing transformational form
FYI, this is much easier from standard form: 442 2 xxy
Find the y-intercept of the following quadratic functions:
Comparing FormsGeneral Form:
Transformational Form:
Standard Form:
Factored Form:
2)HT()VT(VS1
xy
khxay 2)(
c bx axy 2
21VS rxrxy coming soon…
not coming soon…
What are you good for?Transformational form is good for:• finding the vertex
(HT, VT)• finding the range
{y ≤ or ≥ VT, y R}• finding the axis of symmetry
x = HT• finding the max/min value
VT• getting the mapping rule
(x, y) → (x + HT, VSy + VT)• graphing the function• getting the equation from the graph
Drawbacks:• Functions are usually
written in the f(x) notation
2)HT()VT(VS1
xy
What are you good for?General form is good for:• finding the y-intercept
(0, c)• finding the x-intercepts
…coming soon….• finding the vertex
…coming soon….• graphing the parabola
…coming soon…• finding the roots
…coming soon…
c bx axy 2
Drawbacks:• Some work is involved to
obtain this form when only graph, or only transformations are given
What are you good for?
Standard form is good for:• nothing
khxay 2)(
What are you good for?
Factored form is good for:• finding roots
r1 and r2
21VS rxrxy
Drawbacks:• Not all quadratic functions
can be factored
Going Between Forms
general → transformational• 4 steps: includes completing
the square• …coming soon….
transformational → general• 2 steps: FOIL then solve for x
2)1(321
xy 142 2 x – – x y 4 steps
2 steps
Transformational to General Form
FOILirstutsidensideast
2)3( x
)3)(3( xx
2x x3 x3 9
96)3( 22 xxx
Ex. Put the following into general form: 2)4()3(21
xy
)4)(4()3(21
xxy
1644)3(21 2 xxxy
)168(2)3(212 2
xxy
321623 2 xxy
29162 2 xxy
Transformational to General Form
168)3(21 2 xxy
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