Chapter 2: Review – Quadratic Function
V. J. Motto
M110 Modeling with Elementary Functions
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Graphing a Quadratic
Consider the quadratic is y = x2.
Three points will almost certainly
not be enough points for
graphing a quadratic, at least
not until you are very
experienced.
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General form
y = ax2 + bx + c.
The general form of a
quadratic is
y = ax2 + bx + c.
For graphing, the
leading coefficient
"a" indicates how
"fat" or how "skinny"
the parabola will be.
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Which way is up?
There is a simple, if
slightly “silly", way to
remember the
difference between
right-side-up
parabolas and
upside-down
parabolas
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Form y = a(x – h)2 + k
If the quadratic is written in the form y = a(x – h)2 + k, then the vertex
is the point (h, k).
 This makes sense, if you think about it. The squared part is
always positive (for a right-side-up parabola), unless it's zero. So
you'll always have that fixed value k, and then you'll always
adding something to it to make y bigger, unless of course the
squared part is zero. So the smallest y can possibly be is y = k,
and this smallest value will happen when the squared part, x – h,
equals zero. And the squared part is zero when x – h = 0, or
when x = h.
 The same reasoning works, with k being the largest value and
the squared part always subtracting from it, for upside-down
parabolas.
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Let’s Convert some
Place the quadratic in vertex-form:
y = x2 – 6x + 7
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Vertex Formula
An alternate way of finding the vertex (the
maximum or minimum points include:
b 4ac  b2
( ,
)
2a
4a
The formula:
 Use your calculator

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Let’s use the calculator
We will use the calculator to sketch the
graph of y = x2 – 6x + 7 and find the
vertex.
From looking at the graph we see that we
should give the following:
Vertex (maximum, minimum)
 y-intercept
 x-intercepts or zeros

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Finding Zeros by Factoring
You should recall from you study of
Elementary Algebra Foundations the
Zero Product Property. It stated that if
ab = 0
then either a = 0 or b = 0. We can use
this property to solve quadratic equations
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Example 1:
Solve by Factoring
x2  9
x2  9  0
( x  3)( x  3)  0
x  3  0 or x  3  0
x  3
or
x 3
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Subtract 9 from both sides.
Factor.
Zero Product Property
Solve
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Square Root Property
If b is a real number and if a2 = b, then
a2  b
|a| b
a b
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Example 2:
Use the square root property to solve.
x  50
2
x 2  50
| x |  50
x   50
x  5 2
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Example 3:
Use the square root property to solve.
( x  1)2  12
( x  1)2  12
| x  1 |  12
x  1   12
x  1  2 3
x  1  2 3
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Perfect Square Trinomials
Perfect Square Trinomial
Factored Form
x 2  8 x  16
( x  4)2
x 2  6x  9
( x  3)2
9
x  3x 
4
3 2
(x  )
2
2
Notice that for each perfect square trinomial, the constant
term of the trinomial is the square of half the coefficient
of the x-term provided That the coefficient of the squared
term is 1.
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Perfect Square Trinomials
The process of writing a quadratic equation so that
one side Is a perfect square trinomial is called
completing the square.
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Example 4:
Solve by completing the square
x  6x  2  0
2
x  6x
2
2
x  6x  9  2  9
2
x  6 x  9  11
2
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Example 4 (continued)
( x  3)  11
2
( x  3)  11
2
| x  3 |  11
x  3   11
x  3  11 or
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x  3  11
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Example 5:
2x 2  8 x  3  0
3
x  4x   0
2
3
2
x  4x  
2
4 2
So ( )  4.
2
2
x 2  4x  4  
3
4
2
5
( x  2) 
2
2
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Example 5 (continued)
5
( x  2) 
2
2
5
( x  2) 
2
2
10
2
| x 2 | 
x 2 
10
2
10
2
4  10
x
2
x  2
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The Quadratic Formula
Only a few quadratic equations can be
solved by factoring. However, any
quadratic equation can be solved by
completing the square.
Since the same sequence of steps is
repeated each time we completed the
square, let’s generalize the solution into
a formula!
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The Development --- 1
We start with the general quadratic equation:
ax 2  bx  c  0
with a, b and c real numbers and a ≠ 0.
b
c
x  x 0
a
a
b
c
2
x  x
a
a
2
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Divide both sides by a.
Subtract
sides.
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c
a
from both
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The Development -- 2
x2 
b
c
x
a
a
Using the coefficient of the x term, we find the expression
that will complete the square.
2
1 b  b
b2
 b 

and    2


2  a  2a
4a
 2a 
We add this expression to both sides of the equation.
2
2
b
b
c
b
x2  x  2    2
a
4a
a 4a
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The Development --- 3
2
b
b
x2  x  2
a
4a
2
b
b
x2  x  2
a
4a
2
b
b
x2  x  2
a
4a
2
b
b
x2  x  2
a
4a
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c b2
  2
a 4a
c 4a b 2
 
 2
a 4a 4a
4ac b 2

 2
2
4a
4a
b 2  4ac

4a 2
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Find a common denominator
On the right side.
Do the multiplication.
Simplify the right side.
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The Development --- 4
2
b 
b 2  4ac

 x  2a   4a 2


2
b 

 x  2a  


b
x

2a
Factor the perfect square
trinomial on the left
Take the square root of
both sides of the equation
b 2  4ac
4a 2
b 2  4ac
2a
Simplify the radicals
b
b 2  4ac
x

2a
2a
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Solve the absolute value
equation.
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The Development --- 5
b
b 2  4ac
x

2a
2a
b
b 2  4ac
x

2a
2a
Solve for x.
b  b 2  4ac
x
2a
Simplify the expression.
This equation identifies the solutions of the general
Quadratic equation in standard form.
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Comments
b2-4ac is called the discriminant.
If b2-4ac ≥ 0 the solutions are real.
 If b2-4ac < 0, the solution are not real.
 If b2-4ac is perfect square root, the solutions
are integers! (And you could have used
factoring to find the solution.)

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Example 1
2
2
x
 4x  3
Solve
First, write the equation in standard form.
2x 2  4 x  3  0
Identify the coefficients of the quadratic equation
and calculate the value of the discriminant.
a  2, b  4, and c  3
So b2  4ac  (4)2  4(2)(3)  16  24  40
What does this value tell you about the solution?
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Example 1 (continued)
Substitute the values into the quadratic formula.
b  b2  4ac (4)  4  10 4  2 10
x


2a
22
4
The solutions are:
2  10
x
2
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or
2  10
x
2
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Using the calculator
We will use the calculator to find the
zeroes for y = x2 – 6x + 7.
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Group Work
Sketch the graph for identifying all the important points:
y  5x  3x 2  3
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