The powers of General Form
ProbeBelow are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a) Find and correct the odd-one-out.b) Name the 5 forms
x y-1 00 51 82 93 8 4 55 0
2)2()9( xy
542 xxy
)9,2(),( yxyx
ProbeBelow are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a) Find and correct the odd-one-out.b) Name the 5 forms
graphical (parabola)
equation in general form
equation in transformational form
table of values
mapping rule
x y-1 00 51 82 93 8 4 55 0
)9,2(),( yxyx
2)2()9( xy
542 xxy
)9,2(),( yxyx
General Form toTransformational Form
Let’s look at an example.We will take the quadratic function given in general form:
y = 2x2 + 12x – 4and turn it into transformational form:
2)HT()VT(VS1
xy
STEP 1:Divide every term by a.
2621
4122
2
2
xxy
xxy
We know that in transformational form the coefficient of ‘x’ is 1
1
Let’s look at an example.We will take the quadratic function given in general form:
y = 2x2 + 12x – 4and turn it into transformational form:
STEP 2:Move the non-x term over
xxy
xxy
xxy
6221
2621
4122
2
2
2
General Form toTransformational Form
2)HT()VT(VS1
xy
Completing the Squarexxy 62
21 2
x2 x x x x x x
We’re getting close, but the right-hand side of the equation needs to be a perfect square: (x – HT)2
So far we have: x2 + 6xThis looks like:
Completing the Square
x2 x x xx xx Oops…
The equivalent to making x2 + 6x a perfect square is to arrange these tiles into a square shape...
x2 x x x x x xxx 62
Completing the Square
Try again…
x2 x x x x x x
x2 x x x
x
xx
Closer….We just need tocomplete this square…
We were missingnine 1×1 squares.
xx 62
Completing the Square
This looks promising:We have a square with length and width dimensions of (x + 3).We just had to add 9, or 32.
x2 x x x
x
xx
---------- x + 3 --------------------- x + 3
-----------222 )3(36 xxx But how can we justify
this “just add 9”?Add it to both sides of the equation!
But why 32?It’s half of the coefficient of x, squared.
962 xx
Back to our example.We will take the quadratic function given in general form:
y = 2x2 + 12x – 4and turn it into transformational form:
STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)
2
222
2
2
2
)3(1121
363221
6221
2621
4122
xy
xxy
xxy
xxy
xxy
General Form toTransformational Form
2)HT()VT(VS1
xy
Back to our example.We will take the quadratic function given in general form:
y = 2x2 + 12x – 4and turn it into transformational form:
STEP 4:Factor out the 1/a term on the left hand side
2
2
2
2
2
2
)3(2221
)3(1121
969221
6221
2621
4122
xy
xy
xxy
xxy
xxy
xxy
General Form toTransformational Form
2)HT()VT(VS1
xy
Back to our example.We will take the quadratic function given in general form:
y = 2x2 + 12x – 4and turn it into transformational form:
General Form toTransformational Form
Now we know that its VS = 2, HT = –3, and VT = –22 From this we know its vertex, range, axis of symmetry, etc
2)3()22(21
xy
Therefore, the transformational form of the quadratic functiony = 2x2 + 12x – 4 is:
vertex is (–3, –22)
range is{y≥ –22, y R}
axis of symmetry isx = –3
2)HT()VT(VS1
xy
Practice
1. Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.
2. Complete the square to find the range of the functiony = –x2 – 5x +1?
3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.
Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.
1421 2 xxy
xxy 4121 2
444121 2 xxy
2)2(321
xy
2)2()6(21
xy Vertex: (2, –6)
Practice: Solutions
2. Complete the square to find the range of the functiony = –x2 – 5x +1?
152 xxy
xxy 51 2 222 5.255.21 xxy
2)5.2(25.7 xy2)5.2()25.7( xy
Practice: Solutions
range is{y ≤ 7.25, y R}
3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.
262 2 xxyxxy 622 2
96922 2 xxy2)3(112 xy
2)3(2112
xy
Practice: Solutions
A shortcut…eventuallyBut completing the square is a lot of work…. Let’s find a shortcut to finding vertex ect. by completing the
square just ONE more time, but with the general equation
cbxaxy 2642 2 xxy
STEP 1:Divide every term by a.
STEP 2:Move the non-x term over
STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)
STEP 4:Factor out the 1/a term on the left hand side
642 2 xxy
3221 2 xxy
A shortcut…eventuallySTEP 1:Divide every term by a.
STEP 2:Move the non-x term over
STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)
STEP 4:Factor out the 1/a term on the left hand side
xxy 2321 2
222 121321
xxy
211321
xy
21421
xy
21821
xy
cbxaxy 2
acx
abxy
a 21
xabx
acy
a 21
22
2
221
abx
abx
ab
acy
a22
221
abx
ab
acy
a22
241
abx
abcy
a
Here’s the shortcut
This may look complicated, but is VERY helpful…from this we can see that the HT for any general quadratic is
1)2(2)4(
2
x
x
abx
For example: What is the axis of symmetry of the function y = –2x2 + 4x + 6?To complete the square on this takes time. But…
ab2
We now see that ANY general quadratic equationy = ax2 + bx + c can be written as 22
241
abx
abcy
a
Here’s the shortcut
For example: What is the vertex of the function y = –2x2 + 4x + 6?To complete the square on this takes time. But…
abc4
2
We now see that ANY general quadratic equationy = ax2 + bx + c can be written as 22
241
abx
abcy
aThis may look complicated, but is VERY helpful…from this we can see that the VT for any general quadratic is
8)2(6
)2(4)4(6
42
2
yy
y
abcy
vertex (1, 8)
1)2(2)4(
2
x
x
abx
This may look complicated, but is VERY helpful…from this we can see that the VS for any general quadratic is a.
Here’s the shortcut
For example: Graph the function y = –2x2 + 4x + 6.To complete the square on thistakes time. But…
We now see that ANY general quadratic equationy = ax2 + bx + c can be written as 22
241
abx
abcy
a
Since the VS = –2 we can graph from the vertex (1, 8):Over 1 down 2Over 2 down 8Over 3 down 18
Shortcut within a shortcutInstead of memorizing the formula for the y-coordinate of the vertex (VT):
we can calculate it using the general form of the equation and thex-coordinate of the vertex (HT):
For example: What is the maximum value of the function y = –2x2 + 4x + 6?
abx
2
8642
6)1(4)1(2 2
yyy
abcy4
2
1)2(2)4(
2
x
x
abx
Max value is y = 8.
General form shortcuts: Practice
47
21
41a) 2 xxy 42243b) 2 xx=y
For the following quadratic functions, find the vertex, sketch its parabola, give its axis of symmetry, give its range, and write in transformational form, all WITHOUT completing the square.
General form shortcuts: Practice Solutions4
721
41a) 2 xxy
248
47
42
41
47
21
41
47)1(
21)1(
41 2
y
y
y
y
y
12121
412
21
2
x
abx
vertex (1, –2), VS = 1/4
axis of symmetry: x = 1range: {y ≥ 2, y R} 2124
:form tionaltransforma
xy
General form shortcuts: PracticeSolutions42243b) 2 xx=y
6429648
42)4(24)4(3 2
yyy
46243224
2
x
abx
vertex (–4, –6), VS = 3
axis of symmetry: x = –4range: {y ≥ –6, y R} 246
31
form tionaltransforma
x=y
Given a quadratic function in general form y = ax2 + bx + c and a value for y, it is not a straightforward task to find the corresponding x-value(s) because we cannot easily isolate x when it appears in 2 different forms: x and x2.
Subbing in the required y-value to a quadratic function, and bringing all the terms to one side of the equation yields a quadratic equation.
A quadratic equation in general form looks like this:ax2 + bx + c = 0 where a ≠ 0.
Solving a quadratic equation for x is also calledfinding its roots or finding its zeros.
Solving for x given y using general form
Example 1:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = –12.
Solution:−12 = x2 – 2x – 15 0 = x2 – 2x – 3
Although the quadratic function had coefficients a = 1, b = –2, and c = –15,subbing in −12 for y gives the quadratic equation with coefficientsa = 1, b = –2, and c = –3
NOTE: the c value of the quadratic equation might be different from the c value of the quadratic function. See the following example.
Solving for x given y using general form
NOTE: the c value of the quadratic equation is the SAME as the c value of the quadratic function when we are using y = 0. See the following example.
Example 2:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = 0.
Solution:0 = x2 – 2x – 15
Both the quadratic function and the resulting quadratic equation have coefficients a = 1, b = –2, and c = –15.
While this is only true when we use y = 0, this is a very common case, since we are often interested in the x-intercepts of a quadratic function.
Solving for x given y using general form
Soon we will learn how to do this directly from general form…. Last class we learned how to solve this if the function had been given in transformational form…
So lets try that!
Example 3:Find the x-intercepts of the quadratic function y = x2 – 2x – 15
Solving for x-intercepts using general form
Solution:0 = x2 – 2x – 15…but now what?
2)1(16)0( x2)1(16 x
)1(16 x
14 x3
14
x
x5
14
x
x
Therefore the x-intercepts of the parabola are (5, 0) and (–3, 0), and the roots of the function are 5, and –3.
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3: Simplify left-hand side
STEP 4: Take the squarroot of both sides
STEP 5: Isolate x in both equations
2
2
2
2
)1(16
12115
215
152
xyxxyxxyxxy
Solving for x-intercepts using general form
Find the roots of the following quadratics:
11441c) 2 xxy
442d) 2 xxy
xxy 63a) 2
6416b) 2 xxy
Solving for x-intercepts using general formPractice
A shortcut…eventuallyBut this is a lot of work…. Let’s find a shortcut to finding x-intercepts by doing this just
ONE more time, but with the general equation.
cbxaxy 21522 xxy
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3: Simplify left-hand side
STEP 4: Take the square-root of both sides
STEP 5: Isolate x in both equations
A shortcut…eventuallycbxaxy 2
1522 xxy2)1(16 xy
22
221
abx
ab
acy
a22
22)0(1
abx
ab
ac
a
2)1(16)0( x
2
2
2
24
abx
ab
ac
2
2
2
244)(4
abx
ab
aaca
2)1(16 x
2
2
2
2 2444
abx
ab
aac
2
2
2
244
abx
aacb
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3: Simplify left-hand side
A shortcut…eventually2)1(16 x
STEP 4: Take the square-root of both sides
STEP 5: Isolate x in both equations
2
2
2
244
abx
aacb
)1(16 x14 x
abx
aacb
2442
2
abx
aacb
2242
xa
acbab
2
42
2
aacbbx
242
314
xx
514
xx
This is the shortcut, the Quadratic Root Formula!
Back to our example:Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15
Solution:Instead of those 5 steps, let’s apply the Quadratic Root Formula.
2822
6422
6042
12151422
24
2
2
x
x
x
x
aacbbx
52
82
x
x
32
82
x
x
Solving for x-intercepts using general formThe short-cut
The quadratic root formula can be used to find x-values other than the x-intercepts
Solving for x given y using general form
For the function f(x) = x2 – 2x – 15, find the values of x when y = −12:
2422
1622
1242
1231422
24
2
2
x
x
x
x
aacbbx
32
42
x
x
12
42
x
x
Solution:–12 = x2 – 2x – 150 = x2 – 2x – 3.
Now use the quadratic root formua witha = 1, b = –2 and c = –3
Find the roots of the following quadratics:
11441c) 2 xxy
442d) 2 xxy
xxy 63a) 2
6416b) 2 xxy
Solving for x-intercepts using general formPractice
Finding roots from General Form: Practice Solutions
0 let intercept,- for
63a) 2
yx
xxy
0 let intercept,- for
6416b) 2
yx
xxy
2,06666366
)3(2)0)(3(4)6()6(
6302
2
xx
x
x
x
xx
82
016
)1(2)64)(1(4)16()16(
641602
2
x
x
x
xx
Finding roots from General Form: Practice Solutions
0 let intercept,- for
11441c) 2
yx
xxy
0 let intercept,- for
442d) 2
yx
xxy
52821
11164
412
11414)4()4(
114410
2
2
x
x
x
xx
114
1444
1644
32164
)2(2)4)(2(4)4()4(
44202
2
x
x
x
x
x
xx
Find the x-intercepts of the following quadratic functions:
Finding roots from General Form: Practice Solutions
0,2a) xx
8b) x
528c) x
11d) x
Answers:
11441c) 2 xxy
442d) 2 xxy
xxy 63a) 2
6416b) 2 xxy
Yet another method of finding roots: Factoring
As slick as the quadratic root formula is for finding roots, sometimes factoring is even faster!
The zero property:If (s)(t) = 0 then s = 0, or t = 0, or both.(This only works when the product is 0.)
So if we can get the quadratic function in the form:
y = (x – r1)(x – r2), (called factored form)
then when y = 0 we know that:(x – r1) = 0, or (x – r2) = 0, or both.
In other words, x = r1 or x = r2 or both.
Answer when the numbers are added
Factoring to find the roots
We need to factor x2 – 2x – 15.We need 2 numbers whose product is –15 and whose sum is –2
Answer when the numbers are
multiplied
___ × ___ = –15___ + ___ = – 2-5-5
33
Ex. Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15
(There is never more than one pair of numbers that work!)
So the factored form off(x) = x2 – 2x – 15 isf(x) = (x – 5)(x + 3)
So when looking for roots, 0 = (x – 5)(x + 3) and x = 5 or x = –3
So the x-intercepts of the graph are (5, 0) and (–3, 0)
FactoringWhen factoring a quadratic equation where a = 1,find two numbers that multiply to give c and add to give b.
Ex. Find the roots ofy = x2 + 10x – 24by factoring.
24102 xxy ___ x ___ = -24
___ + ____ = 10)2)(12( xxy
)12(0 x or )2(0 x12x 2x
By the way:(0) = x2 + 10x – 24 24 = x2 + 10x24 + 25 = x2 + 10x + 2549 = (x + 5)2
±7 = x + 5x = –5 + 7 or x = –5 – 7x = 2 or x = –12
12or2224or
24
21410
219610
)1(2)24)(1(41010 2
xx
xx
x
x
x
)2)(12(0 xx
12
12
-2
-2
Practice1. Solve the following equations by factoring.
a) 0 = x2 + 2x +1
b) 0 = x2 + 5x +4
c) 0 = x2 + 2x – 24
d) 0 = x2 – 25
2. Find the x- and y- intercepts of the following quadratics. a) f(x) = 2x2 +3x + 1
b) f(x) = 3x2 – 7x + 2
c) f(x) = x2 – 5x -14
d) y = 2(x – 4)2 – 32
Answers 1a. x = –1 b. x = –4 or –1 c. x = –6 or 4 d. x = –5 and 5
2a. (–1, 0) (–0.5, 0) (0, 1) b. (1/3, 0) (2, 0) (0, 2) c. (–2, 0) (7, 0) (0, –14) d. (0, 0) (8, 0)
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