Download Chapter Five: Quadratic Functions Section One: Introduction to

Chapter Five: Quadratic Functions
Section One: Introduction to Quadratic Functions
Back in chapter one we looked at linear equations of the form y  mx  b . We will now examine a
functions that is a little more complex. A quadratic function is any function of the form
y  ax 2  bx  c , where a is not zero.
EX1: Show that the following is a quadratic function and identify a, b, and c.
y  (3x  4)(2 x  3)
The graph of a quadratic function is called a parabola.
All parabolas are symmetrical about an axis of symmetry. This axis always passes through the extreme
minimum (or maximum) of the graph which is referred to as the vertex. We can determine whether the
vertex is a maximum or minimum by looking at the “a” value.
y  2 x 2  3x  4
y  2 x 2  3 x  4
Notice that when the value of “a” is positive, the parabola cups up and the vertex is the minimum point.
When “a” is negative the parabola cups down and the vertex is the maximum point.
EX2: In the following, tell whether or not the function is quadratic. If yes, identify the values of a, b, and
c. Then tell whether the parabola cups up or down and if the vertex is a maximum or minimum. Finally,
approximate the value of the vertex by using a graph.
a. y  2 x 2  2 x
b.
c.
y  2  5x  2 x 2
y  3( x  4)(2 x  1)
Section Two: Introduction to Solving Quadratic Equations
We need to first review the concept of the square root. When we are asked to simplify 9 we know
that we get 3. This is referred to as the principal square root (the positive square root). However, the
number nine actually has two square roots, 3 and -3. This is because 32  9 and  3  9 . We need to
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remember this fact when solving quadratic equations.
When solving a simple quadratic equation (the “b” term is missing), we can simply take the square root
to get rid of an exponent of two. This is because squares and square roots are inverse operations just
like addition and subtraction.
Note: Since we know that we must consider both the positive and negative square root, we sometimes
use the symbol  .
EX1: Solve the equation x 2  25
We can use basic properties of radicals to simplify some square roots:
1.
ab  a  b
2.
a
a

b
b
EX2: Simplify the following radicals using the previous rules
a.
b.
50
300
c.
15
16
d.
25
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Solving Basic Quadratic Equations
1. Isolate the squared term
2. Take the 
of each side
3. Solve the resulting linear equation(s)
EX3: Solve the following equations
a. 7 x 2  23  103
b.
64  x  5   289
c.
36  x  3  12  38
2
2
EX4: A humanitarian group provides assistance to people who need help in obtaining resources. The
group has decided to send a shipment of food by helicopter to some families living in a remote area. The
helicopter, which is 85 feet above the designated drop-off area, releases a box containing food. The
height of the box can be modeled by h  t   16t 2  85 , where t is the time in seconds after it is
released. After how many seconds will the box land on the ground?
In any right triangle, the length of a missing side can be found by using the Pythagorean Theorem:
a 2  b 2  c 2 , where a and b are the lengths of the legs of the triangle and c is the length of the
hypotenuse.
EX5: Find the length of the missing legs of the following right triangles
a.
b.
EX6: A 75-foot tall tower has a support wire that is 106 foot long. We want to add a second support
wire. The distance to the second wire is to be half the distance to the already existing wire. How long
should our new support wire be?
Section Three: Factoring Quadratic Expressions
Factoring is simply the opposite of multiplication. It is the process of breaking a product back into
factors.
2  3  6 (Multiplying) 6  2  3 (Factoring)
2( x  3)  2 x  6 (Multiplying) 2x  6  2(x  3) (Factoring)
We should follow these steps when trying to factor an expression
1. Look for any common factors that can be pulled out of every term
2. Do we have any special cases
a. Difference of squares
b. Difference or sum of cubes
3. Do we have a trinomial with an “a” that is 1
4. Do we have a trinomial with an “a” that is not 1
EX1: Factor out the GCF
a. 27c 2  18c
b. 5 z (2 z  1)  2(2 z  1)
Below are the formulas for the special cases listed above. We need to pay close attention to difference
of squares because we see this case occur a lot.
Difference of Squares: a 2  b2  (a  b)(a  b)
Difference of Cubes: a3  b3  (a  b)(a 2  ab  b2 )
Sum of Cubes: a3  b3  (a  b)(a 2  ab  b 2 )
EX2: Factor the following binomials
a. x 2  9
b. 16 x 2  25 y 2
c.
d.
x 4  16
x3  27
If we have a trinomial of the following form, x 2  bx  c , we can factor by asking these two questions:
1. What two numbers add to give us b?
2. What same two numbers multiply to give us c?
EX3: Factor the following trinomials
a. x 2  12 x  27
b. x 2  7 x  30
c. x 2  15 x  54
d. x 2  15 x  36
If “a” does not equal 1, we have to use the following procedure
EX4: Factor the following trinomials
a. 6 x 2  11x  3
b. 3 x 2  5 x  2
We have a shortcut for factoring special trinomials called “perfect-square trinomials.”
EX4: Factor the following trinomials
a. x 2  12 x  36
b. x 2  8 x  16
Note: In the previous examples,  half of b   c . This tells us we have a perfect-square trinomial.
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If things are multiplied and the product is zero, we know that one of the factors must be zero. This is
called the zero product property.
If a  b  c  d  0 , then we know that one of the four factors must be zero.
EX5: Use factoring and the zero product property to find the zeros of the quadratic equations
a. f ( x)  x 2  4 x  21
b.
g ( x)  x 2  3x  40
h( x )  4 x 2  8 x  3
d. j ( x)  2 x 2  6 x
c.
Section Four: Completing the Square
Another method we can use to solve quadratic equations is completing the square. Not only will we use
it to solve equations, but we will also need this information in the next section and later on in chapter
nine when dealing with conic sections.
Before we begin completing the square we need to review perfect square trinomials. In a trinomial, if
 half of b 
2
 c then it is a perfect square. This is our goal!
EX1: What value would complete the square
a. x 2  8 x ____
b.
x 2  3x ____
Steps for Completing the Square
1. Write your equation in standard form
2. Move the constant term to the opposite side
3. If “a” is not one, factor “a” out of the two terms
4. Complete the square (add the same value to both sides)
5. Factor the completed square
6. Simplify the other side
7. Solve the resulting equation
EX2: Solve by completing the square. Check with a graph.
a. x 2  18 x  40  0
b. x 2  10 x  24
c. 3x 2  6 x  5
We can use completing the square to change a parabolas equation from standard form to vertex form. If
we review transformations, we know how the parabola is different from its parent function f ( x)  x 2 .
Standard Form
f ( x)  2 x 2  12 x  14

Vertex Form
 f ( x)  2( x  3) 2  4
From the previous example, we see that our graph has been flipped across the x-axis, compressed
horizontally by a factor of 12 , translated right 3, and translated up 4. Notice that in f ( x)  x 2 our
vertex is always at (0, 0) . Since we translated our graph, we can easily identify our new vertex of (3, 4) .
The ease of finding the coordinates of our vertex is why we call this new form vertex form.
EX3: Write the following in vertex form. Describe all transformations. Give the coordinates of the vertex
and the equation of the axis of symmetry.
a. g ( x)  2 x 2  12 x  13
b.
f ( x)  3x 2  9 x  2
How could we find the x-intercepts and y-intercepts?
Section Five: The Quadratic Formula
The basic form of quadratic equations is ax 2  bx  c  0 . If we use completing the square to solve this
for x we get x 
b  b2  4ac
. This is called the quadratic formula. This formula is a shortcut for
2a
solving any quadratic equation. We simply identify our a, b, and c and then plug them into the formula.
EX1: Solve the following equations
a. x 2  5 x  14  0
b. 5 x 2  1  7 x
c. 4 x 2  8 x  3
EX2: We want to build a patio around our house as seen below. We want the area of the patio to be 600
square feet. What value of x would give us this area?
Using the quadratic formula and the midpoint formula gives us yet another way to find the vertex of a
parabola: x 
b
. This formula gives us the x-value of the vertex. We then simply plug this x back into
2a
the original to find y.
EX3: Find the vertex of the following parabola
a. g ( x)  2 x 2  12 x  13
b.
f ( x)  3x 2  9 x  2
Section Six: Quadratic Equations and Complex Numbers
As we have seen, quadratic equations could have 2, 1, or even 0 roots. By using what we know about
square roots we kind determine how many solutions a quadratic equation has by looking simply at the
discriminant in the quadratic formula (the stuff under the square root).
If b 2  4ac  0 , the equation will have two real solutions
If b 2  4ac  0 , the equation will have one real solution
If b 2  4ac  0 , the equation will have no real solutions
EX1: Determine the number of solutions by using the discriminant. Check with a graph.
a. f ( x)  2 x 2  4 x  1
b.
c.
f ( x)  2 x 2  4 x  2
f ( x)  2 x 2  4 x  3
Even though the equation in example c has no real solutions, it does have two complex solutions. To
discuss complex numbers we will first define imaginary numbers. The number i is an imaginary number
such that i  1 . All numbers that we have discussed thus far have been real numbers. However,
there is another realm of numbers that exist. The collection of all real and imaginary numbers are called
the complex numbers.
Since we now know that we can write the
1 as i , we can use this to clean up non-real solutions.
EX2: Write the following in terms of i .
a.
16
b.
8
200
c.
EX3: Solve the quadratic equation. Leave answers in terms of i .
6 x 2  3x  1  0
A standard form of a complex number is a  bi , where a is referred to as the real part and b is the
imaginary part. Two complex numbers are equal if and only if both the real and the imaginary parts are
equal.
EX4: Find x and y such that 3x  4iy  21  16i
Since we know that i  1 , we can find multiples of i .
i  1
i 2  i  i  1  1  1
i 3  i 2  i  1 i  i
i 4  i 2  i 2   1   1  1
i 5  i 4  i  1 i  i
i 6  i 4  i 2  1   1  1
And the pattern continues …
We can use the fact that i 4  1 to find higher powers of i .
EX5: Find the value of the following
a. i13
b. i 27
c. i164
We can add, subtract, and multiply complex numbers by using previous knowledge of binomials.
EX6: Find the sum or difference
a.  10  6i   8  i 
b.
 9  2i   3  4i 
EX7: Multiply
a.  5  3i  6  4i 
b.
 2  4i 
2
Math etiquette says that an imaginary number (and square roots) should not be left on the bottom of a
fraction. We can rid denominators of i's by multiplying by a conjugate. We call this process rationalizing
the denominator.
EX8: Rationalize the denominator:
3  4i
2  5i
The coordinate plane we are used to dealing with only works for real numbers. Therefore, complex
numbers cannot be graphed on the Cartesian plane. We graph complex numbers on a complex plane.
This plane is made up of a horizontal real axis and a vertical imaginary axis. We follow the same process
as with graphing real points.
EX9: Plot the following complex numbers on a complex plane
a. 5  3i
b. 6  4i
The absolute value of a real number is the distance that number is from zero on a number line. The
absolute value of a complex number is the distance that number is from the origin in the complex plane.
We can use the Pythagorean theorem (or distance formula) to find the absolute value of a complex
number.
EX10: Find the following absolute values
a. 5  3i
b.
6  4i
Section Seven: Curve Fitting with Quadratic Models
Most of the time given three points that do not form a line, we can fit a parabola through the three
points perfectly. We follow these steps:
1. Plug the three points into the standard form of the equation: y  ax 2  bx  c
2. You should now have three equations with three variables: a, b, and c
3. Solve the system for the missing variables
4. Plug these values back into y  ax 2  bx  c
EX1: Find the quadratic function whose graph contains the points  3,16 ,  2, 6  , and 1, 4
EX2: Find the quadratic function that models the following sequence: 2, 6, 12, 20…
Sometimes a parabola won’t fit the data perfectly but is a close approximation. This topic is very similar
to the line-of-best fit from chapter one. We can find a quadratic regression equation to closely match
the data.
EX3: Make a scatter plot for stopping distance vs. speed. Find the quadratic regression that closely
resembles the data. Approximate the stopping distance if you are traveling 100 mph.
Speed (mph) Stopping Distance (ft)
20
44
30
83
40
132
50
193
60
264
70
347
Section Eight: Solving Quadratic Inequalities
The method for solving quadratic inequalities is identical to that of solving absolute value inequalities
from Chapter one, section eight.
Solving Quadratic Inequalities
1. Find and plot critical values on a number line
a. Solve the corresponding equation to find these
2. Pick a number from each region on the number line
3. Test these numbers by plugging them into the original inequality
4. Graph accordingly
EX1: Solve the following inequalities
a. x 2  3 x  4  0
b. x 2  8 x  12  0
c.
3  x  1  0
d.
 x  4
2
2
0
Quadratic inequalities with two variables are graphed with the same rules as linear inequalities with two
variables:
1. Get the corresponding equation in y  ax 2  bx  c form and graph
2. If there is an “or equal to” line, the line will be solid. If there isn’t, the line is dotted
3. For “greater thans” shade above the parabola. For “less thans” shade below the parabola.
EX2: Graph the following quadratic inequalities
a.
y   x  1  5
b.
y    x  4  7
2
2