How are you progressing in math these days?
Are you seeing the 'big picture' re. quadratic relations?
wrt the general form: y = f(x) = a*x*x + b*x + c
1) the graph is a parabola ... And the graph is symmetric about about the line:
x = -b/(2*a)
2) it opens up with a min extreme value if a > 0, i.e. if a is pos.
3) it opens down with a max extreme value if a < 0, i.e. if a is neg.
(Recall that here, a != 0, i.e. a is NOT zero, since if a is zero, we no longer have an x**2 term.)
4) it has a pair of roots (zeros) about the axis of sym-metry, the line:
x = -b/(2*a)
And the two roots here are given by: (-b +- sqrt(b*b - 4*a*c))/(2*a)
Or ... re. the discriminant, D = (b*b - 4*a*c),
in case 1, we take the pos. sqrt(D), so y = 0 at x = (-b + sqrt(D))/(2*a)
in case 2, we use the neg. sqrt(D), so also y = 0 at x = (-b - sqrt(D))/(2*a)
Note! If D = 0, then we say that we have just one ... (or 'two equal roots') ...
i.e. the graph of the quadratic just 'touches' the x-axis.
If D > 0, then we have two *real* roots,
If D < 0, we say that roots are 'imaginary',
i.e. the roots in the case where D is neg. involve the use of the 'imagined number': sqrt(-1)
Note that we obtained this general formula for the roots of: y = a*x*x + b*c + c
by using the 'completing the square' method.
So now, via this general formula, and understanding the significance of D, the discriminant,
we can quickly just sub into the formula and find the roots,
and do not need to use factoring method, or completing the square method,
which were steps on the way to understanding the quadratic's (paired) roots
and how we derive the general formula for the roots.
Note in particular, that if D evaluates to zero,
then that is the case of there being only one root, (and the graph just touches the x-axis),
i.e., the case where y = 0 ONLY at this one value of x = ?
Hint: what is equation for axis of symmetry?
x = ?
Answer:
x = -b/(2*a) is equation of axis of symmetry for all parabolas and
so this x, is also the one value of x if a parabola's a, b, c's are such that
the graph only touches the line.
You can also see this easily from setting D = 0 in the general formula.
Then roots are: (-b +- sqrt(0))/(2*a) = -b/(2*a)
Can you now see that this all makes sense, since the parabola is symmetric about: x = -b/(2*a)
So if the (a,b,c's of a) parabola (are such that it) just touches the x-axis at only one place,
then it must be there, at x = -b/(2*a)
Ok ... let's now ...
Start back at the simplest example of a quadratic relation:
What is graph of:
y = x*x
Then ask:
What changes if we graph:
y = a*x*x where a is pos or neg and the absolute value of a gets bigger than one ... or ... gets smaller than one ?
How does the graph change if we graph:
y = a*x*x + k
?
Then, finally, how does the graph change if we graph:
y = a*(x - h)**2 + k
?
Note that the final form above is the ... 'Vertex form' of a quadratic with the vertex point being: (h, k)
i.e. the extreme value of y = k occurs here at x = h
and is a max if a < 0
or is a min value if a > 0
Note that if a < 0, then for values of x different from h, we are always subtracting a value from k ... So k is the max
And if a > 0, then for values of x different from h, we are always adding a value to k ... So k is the min value.
Note that the 'extreme' value of: y = k
in this vertex form, always occurs at: x = h
The only other form you were given was the root form, where the roots are r1 and r2: y = a*(x - r1)*(x - r2)
And you can expand this now and find ...
y = a(x*x -(r1+r2)*x + r1*r2)
= a*x*x - a*(r1+r2)*x + a*r1*r2
And comparing the relevant coefficients here with the general equation that uses
parameters of a, b, c for a quadratic, we can see that:
a is same ...
and ...
b = -a*(r1+r2)
i.e.
(r1+r2)/2 = -b/(2*a)
which we knew already from symmetry, that the extreme value for y occurs at x = mid point of roots,
and finally ... (comparing coefficients)
c = ?
here c = a*r1*r2
Or alternately:
r1*r2 = c/a
and ...
(r1+r2) = -b/a
So ... back now to practicing completing the square method.
Try this:
y = 3*(2-x)*(1+5*x)
We know here (in advance by inspection) that the roots are: 2 and -1/5 since a factor becomes zero with x taking either one of these values.
Firstly, expand ...
then see what you can do? Can you expand it?
Recall that: 3*(a + b) = 3*a + 3*b (distributive law)
And so also we get:
(a + b) * (c + d) =
a*c + a*d + b*c + b*d
Hint: when you expand it,
3*(2-x)*(1+5*x)
in your first step ...
leave the times 3 factor out front the brackets, where, inside the brackets,
you show the expansion of the other two factors.
So 1st step: y = 3*(2-x)*(1+5*x)
Firstly, expanding we get: y = 3*(2 + 9*x - 5*x*x)
Note: we were probably here starting with: y = 6 + 27*x - 15*x*x
and then our 1st step might be,
(after dividing each term by -3 and rearranging with -3 factor out front all):
y = -3*(5*x*x -9*x -2)
= -3*(5*x*x -9*x) + 6
Now starting with : y = -3*(5*x*x -9*x) + 6
we might next divide the 5 and 9 by 5 and put the 5 factor out front,
so ...
y = -15*(x*x - 9/5*x) + 6
Now what is (1/2 of (-9/5)) squared?
thus: y = -15*(x*x - 9/5*x + 81/100 - 81/100) + 6
= -15*(x - 9/10)**2 + 15*81/100 + 6
And if y is 0, then dviding both sides by -15 gives ...
0 = (x - 9/10)**2 - 81/100 - 6/15
Or ...
(x - 9/10)**2 = 81/100 + 6/15
= (243 + 120)/300
= 363/300
= 121/100
So taking square roots of both sides we get:
x - 9/10 = +- 11/10
Thus ...
y is zero if ...
x = 9/10 + 11/10 = 2
or ...
x =9/10 -11/10= -1/5
And that is probably the hardest completing the square example you are ever likely to get on a test ...
The trick here, on a test, if you need to show steps to completing square method,
would be to firstly use the general formula to quickly find the roots ...
and then, knowing what your final values should be,
that can help you see if all your algebraic manipulating is on track
But can you see now ... why it is a good idea to do all this grunt work, (once),
to develop the general formula for the roots of a quadratic,
by the 'completing the square method' ...
then ...
all we ever need to do after ... is to sub the values of a, b, c into the formula, to find the roots?
What do we mean when we say a number is itself squared?
Recall that for a square room, with sides 4 ft by 4 ft ... the area is 16 sq. ft.
Or the other way around ...
if a perfectly square room had area 16 square feet, what is the length of each side?
Answer: Each side is 4 ft. long.
Recall I am using Python exponent notation of ** ... so ... by using ... 2**3
I mean to indicate: 3 factors of 2
So 4**4 = ?
Four factors of 4 = ?
4*4*4*4 = ?
Same as : 16 * 16
Same as : 2**8 = 256
Recall 4 = 2**2
and so 4**4 = (2**2)**4 = 2**(2*4) = 2**8 = 256
We maybe need to do more practice with exponents ?
What is 3**2 ?
What is 4**2 ?
What is 5**2 ?
What is sqrt(81) ?
What is sqrt(144) ?
What is sqrt(49) ?
Recall ...
I am NOT using 'x' to indicate times ...
since we are using 'x' as an independent variable in all our functions here like y = f(x)
And the Python language ... and all standard computer programming languages ...
they pretty much all use calculator notation to indicate multiplication by using the star. *, i.e. the asterisk symbol.
Note also ... that in all computer programs (and calculators) you can NOT code: x = 3(4+5)
you must instead, always code like this: x = 3*(4+5)
So now memorize the general formula to find the roots of any quadratic ...
For: y = f(x) = a*x**2 + b*x + c
Roots are: (-b +- sqrt(b*b-4*a*c))/(2*a)
Note: we need to take both + and - cases to find both roots if discriminate D = (b**2 - 4*a*c) is not zero.
NOW ... practice more on completing the square method on some easier equations that you can make yourself by using root form:
y = a*(x-r1)*(x-r2)
and sub in easy values for roots r1, r2 and also for a.
Then expand and practice completing the square method of finding the roots from that starting point ... already knowing what the roots each are
If you get stuck ...I'm only a quick e-mail away
The Lord knows all about perfect squares since He created it all perfect.
Bless the Lord oh my soul, and all that is in me, bless His Holy Name!!!
Baruch HaShem!!!
(Bless His Name)
Praise the Lord!!! ... He has promised to never leave or forsake those who place their trust in Him..
Recall also His encouraging words to these ones, that He speaks at the end of Matthew's Gospel ...
ALL power/authority is given to me in heaven and in earth ... go ... I am with you alway, (even) unto the end of the world.
(Matthew 28:16-20)
16 Then the eleven disciples went away into Galilee, into a mountain where Jesus had appointed them.
17 And when they saw him, they worshipped him: but some doubted.
18 And Jesus came and spake unto them, saying, All power is given unto me in heaven and in earth.
19 Go ye therefore, and teach all nations, baptizing them in the name of the Father, and of the Son, and of the Holy Ghost:
20 Teaching them to observe all things whatsoever I have commanded you: and, lo, I am with you always, even unto the end of the world. Amen.
And recall:
But of him are ye in Christ Jesus, who of God is made unto us wisdom, and righteousness, and sanctification, and redemption. (1 Cor. 1:30)
And ...
For he (God) hath made him (Jesus to be) sin for us, who knew no sin; that we might be made the righteousness of God in him (in Jesus.) (2 Cor. 5:21)
