Download Introduction to Quadratics – Summary Guide

Introduction to Quadratics – Summary Guide
CHAPTER THREE:
THE PARABOLA - VOCABULARY
Shape of a Quadratic Relation is known as a parabola (U-shaped
curve in diagram)
x-intercept(s) – location where the curve crosses the x-axis; also
known as: roots or zeroes
y-intercept – location where the curve crosses the y-axis
axis of symmetry (aos) – vertical line which cuts the parabola
into 2 equal halves. Equation starts with x =; it is the xcoordinate of the vertex.
optimal value – y-coordinate of the vertex. Maximum or
minimum value.
vertex – the highest or lowest point of the parabola
MULTIPLYING BINOMIALS - EXPANDING AND SIMPLIFYING (Changing from factored form to standard form)
x  3x  2
 xx  2   3x  2
1.
 x 2  2 x  3x  6
← I took the first term in the first bracket and multiplied it by the entire second
bracket; then I took the second term in the first bracket and multiplied it by the entire
second bracket
← Combine like terms
 x2  x  6
x  52
  x  5 x  5
2.
 x x  5  5 x  5
 x 2  5 x  5 x  25
 x 2  10 x  25
←I wrote out the bracket twice due to the exponent of 2
← I took the first term in the first bracket and multiplied it by the entire second
bracket; then I took the second term in the first bracket and multiplied it by the
entire second bracket
← Combine like terms
2 x  1x  2 
3.
 2 x  2  x  2 
←The coefficient 2 needs to be multiplied into one of the brackets
 2 x x  2   2 x  2 
← I took the first term in the first bracket and multiplied it by the entire second
bracket; then I took the second term in the first bracket and multiplied it by the
entire second bracket
← Combine like terms
 2x2  4x  2x  4
 2x2  2x  4
FACTORING (Changing from standard form to factored form)
Common Factoring
2x2  6x  2
1.


 2 x 2  3x  1
← The greatest common factor (GCF) of this trinomial is 2. I divide out the 2 from each of
the terms in the polynomial. The 2 goes after the equal sign and in an attached bracket is
the remainder from the division of the trinomial by 2.
9 x 4 y 4  12 x 3 y 2  6 x 2 y 3
2.

 3x 2 y 2 3x 2 y 2  4 x  2 y

← GCF of the numbers 9, 12, and 6 is 3. The GCF of the variable parts x4y4, x3y2,
and x2y3 is x2y2. Therefore, the GCF is 3x2y2. I divide out 3x2y2 from each of the
terms in the polynomial. The 3x2y2 goes after the equal sign and in an attached
bracket is the remainder from the division of the trinomial by 3x2y2. When
dividing like variables, subtract the exponents.
NOTE: Every time you approach a question where factoring is required you must always common factor first
Simple Trinomial Factoring (x2 + bx + c, where a = 1)
x2  6x  8
3.
 x  4x  2
← Check for a common factor first. Since this is a trinomial where a = 1, I must determine
two numbers which multiply to 8 (the c-value) and add to 6 (the b-value). These numbers
are +2 and +4. We write two brackets, placing x’s in the front of each bracket and the +2
and +4 in the back of one of the brackets, separately.
HINT: Look for simple trinomial factoring if a = 1 and it is a trinomial.
Decomposition (ax2 + bx + c, where a ≠ 1)
12 x 2  25 x  12


 12 x 2  16 x   9 x  12 
4.
 4 x3 x  4   33 x  4 
 3 x  4 4 x  3
← Check for a common factor first. Since this is a trinomial where a ≠ 1, I must
determine two numbers which multiply to the product of a x c, which is 144 and
add to -25 (the b-value). These numbers are -16 and -9. We copy out the first
term and the last term again however, replace the middle term with -16x and
-9x. Brackets are placed round the first two terms and the last two terms (if the
third term is negative, the negative sign must be included in the back bracket).
Common factor each bracket, and then divide out the common bracket,
grouping together the common factors.
HINT: Look for decomposition if a ≠ 1 and it cannot be common factored out.
Differences of Squares
x 2  64
5.
  x  8 x  8
←This is a binomial where the perfect square terms are separated by a subtraction sign.
After looking for a common factor, take the square root of both terms. In the answer,
there are two different brackets formed, where the square root of the first term goes in
the front part of each bracket and the square root of the second term goes in the back
half of the bracket, with a plus placed in one bracket and a minus in another.
HINT: Look for differences of squares if a binomial, each term is a perfect square, and there is a subtraction sign
separating the terms
Perfect Square Trinomial
25 x 2  40 x  16
6.
 5 x  4 5 x  4 
 5 x  4 
2
← Remember to look for a common factor first. In a perfect square trinomial, the first
and third terms are perfect squares, which when square rooted and multiplied
together and by the number 2 equals the b-value (i.e. 25  5 and 16  4 , 5 x 4 x 2
= 40). Take the square root of the first term and place it in the front of two brackets;
take the square root of the last term and place it in the back of the two brackets; and
place the sign of the b-value in between the two parts. Both brackets will be identical
and therefore, write only one bracket in the final answer raised to the exponent 2.
HINT: Look for perfect square trinomial if a trinomial where the first and last terms are perfect squares and the product
of their square roots times 2 is equal to the b-value.
ZEROS, AXIS OF SYMMETRY, OPTIMAL VALUE, AND VERTEX
Standard Form
y  x2  6 x  8
Factored Form
y  x  4x  2 → Use the factoring techniques mentioned above to convert the standard
Zeros
form equation to factored form
Set each factor equal to zero and solve for the zeros (x-intercepts) by re-arranging for x.
x40
x  4
Axis of Symmetry
AND
x20
x  2
Add the x-intercepts together and divide by 2.
 4   2
2
6
aos 
2
aos  3
aos 
Optimal Value
Replace the x-value from either the standard form or the factored form equation with the axis of
symmetry value from above to determine the y-coordinate of the vertex, aka optimal value.
Sub x = -3 into the equation below:
Sub x = -3 into the equation below:
y  x2  6x  8
y   x  4  x  2 
y  ( 3) 2  6( 3)  8
y  (( 3)  4)(( 3)  2)
y  9  18  8
y  1
Vertex
Remember, the aos value is the x-coordinate of the vertex.
y  (1)( 1)
y  1
The vertex is the highest or lowest point, which is constructed from the axis of symmetry value and
optimal value (aos, optimal value).
(-3, -1)
HELPFUL HINTS WHEN IT COMES TO WORD PROBLEMS:
What is the question?
What is the question asking for?
AND
What math is involved here?
What is the initial value?
This question is asking for the y-intercept.
(i.e. initial height, initial population)
This is equal to the c-value of the standard form equation.
To calculate this value from either a standard form or factored form equation
replace the independent variable (the one that’s after the equal sign) with the
number 0 and complete the math.
When does the ball hit the ground?
These questions are all asking for the zeros (x-intercepts).
What are the break-even points?
You need an equation that is in factored form, then solve for your zeros by
What is the time when the population setting each factor equal to zero and solve for the independent variable. You
is zero?
need to see if both values make sense for the question asked.
What is the time the ball reaches
This question is asking for the axis of symmetry value.
maximum height?
This question needs to mention either the word maximum or minimum (or a
What is the number of items sold to
synonym of these words).
make maximum profit?
From a factored form equation, find the zeros, then add the zeros together and
When does the population reach its
divide by 2 to determine the axis of symmetry value.
maximum?
What is the maximum height the ball
This question is asking for the optimal value.
reaches?
This question needs to mention either the word maximum or minimum (or a
What is the maximum profit?
synonym of these words).
What is the maximum population?
From a factored form equation, find the zeros, then the axis of symmetry, and
then substitute the axis of symmetry value into either the standard or factored
form to find the optimal value.
What is the height of the ball when
This question is asking for the dependent value given a specific independent
time is _____?
value.
What is the profit when ________
Substitute the independent value into the equation and solve for the
items have been sold?
dependent value.
What is the time when the ball
This question is asking for the independent value given a specific dependent
reaches _______ height?
value.
What is the number of items sold to
Substitute the dependent value into a standard form equation, re-arrange
make ________ profit?
equation and set it equal to zero. Change to factored form equation and find
your zeros.