Download Vertex Form of Quadratic Equations

Vertex Form
of Quadratic
Equations
This PowerPoint was selected
for my Algebra1-HP students as
a result of conducting a simple
Google search. The result
contained a small logo entitled
“Solo Math”. It is well prepared
and I apologize to the authors
for revising some sections to suit
the academic needs of my
students. I cannot take full
credit for the good and very
comprehensive content found
here. I hope that it is of use to
the viewer.
General Form
 Quadratic
functions have the standard form
y = ax2 + bx + c


a, b, and c are constants
a ≠ 0 (why?)
 Quadratic
functions graph as parabolas or U’s
Zeros of the Quadratic
 Zeros
are where the function crosses the x-axis. They
are also called roots and solutions.

Where y = 0
 Consider
possible numbers of zeros

None (that are Real): or
two complex/imaginary)
One


Two
Axis of Symmetry
 Parabolas
are symmetric
about a vertical axis
 For y = ax2 + bx + c the axis
of symmetry is at
b
x
2a
 Given
y = 3x2 + 8x
What is the axis of symmetry?
Vertex of the Parabola
 The
vertex is the “point” of the
parabola. It is a turning point.


This is a vertex with a minimum value
A vertex can also be a maximum
 What
is the x-value of the
vertex?
 How
can we find the
y-value?
b
x
2a
 b 
y  f ( x)  f  
 2a 
Vertex of the Parabola
 Given
f(x) = x2 + 2x – 8
 What is the x-value of the vertex?
b 2
x

 1
2a 2 1
 What
is the y-value of the vertex?
f (1)  1  2  9  9
 The
vertex is at (-1, -9)
Vertex of the Parabola
 Given

f(x) = x2 + 2x – 8
Graph shows vertex at (-1, -9)
Your can use your calculator to find the vertex:
minimum or maximum!
Quadratic Forms
 Standard
form
y = ax2 + bx + c
 Vertex form
y = a (x – h)2 + k

Then (h,k) is the vertex
 Given


f(x) = x2 + 2x – 8
Change to vertex form
Hint, use completing the square
Vertex Form
 Changing
to vertex form
y  x  2x  8
2
y  x  2x 
8
2

y  x
Now create a
squared
binomial

2

Add a number to make
a perfect square
trinomial
Subtract the same *
amount to keep it
even.
This gives us the
ordered pair (h, k)
* Instead, you may choose to send the constant over to the other side first
and add the same number to both sides in this step. This is how we learned it.
• What's the pattern?
x
(x + 6)2
+
6
x
x2
6x
+
6
6x
36
• How about these?

x2 + 10x ______
+ 25 
x2 – 14x ______
+ 49 
+ 4
x2 + 4x ______
(x _____
+ 2 )2
(x _____
+ 5 )2
(x _____
– 7 )2
x2 + 12x + 36
Transforming from standard form to vertex form can be
easy…
x2 + 6x + 9  (x + 3)2
62  3
32  9
x2 – 2x + 1
= (x – 1)2
x2 + 8x + 16
= (x + 4)2
x2 + 20x + 100 = (x + 10)2
… but we're not always so lucky
• The following equation requires a bit of work to get it
into vertex form.
16 is added to both sides in
y = x2 + 8x + 10
this problem to complete
-10 + 16= (x2 + 8x + 16)
82  4
42  16
y = (x + 4)2 – 6
the square. [16 could have
been subtracted to maintain
the balance of the equation
if I had not sent the 10 over
to the left.] Then the
constants on the left are
simplified and brought back
to the right side.
The vertex is located at ( –4, –6 ) -Be very careful with the signs.
• Lets do another. This time the x2 term is negative.
– x2 +
y=
12x – 5
y = (– x2 + 12x – 5)
y = – (x2 – 12x + 5)
)
– 5 = – (x2 – 12x
+ 36 – 5 = – (x2 – 12x + 36 )
+ 36
+ 31
– 31 + 31
= – [(x – 6)2
]
= – [(x – 6)2 – 31 ]
y = – (x – 6)2 + 31
The vertex of this parabola is located at ( 6, 31 ).
Un-distribute a negative
so that you can complete
the square

12  2 6
(  6) 2  36
Bring back the constant
and then re-distribute a
negative after
completing the square.
• The vertex is important, but it's not the only important
point on a parabola
y-intercept at (0, 10)
x-intercepts at
(1,0) and (5, 0)
Vertex at (3, -8)
• In addition to telling us where the vertex is located the
vertex form can also help us find the x-intercepts of the
parabola. Just set y = 0, and solve for x.
y = (x + 4)2 – 6
0 = (x + 4)2 – 6
Add 6 to both sides
6 = (x + 4)2
Take square root of both sides
6  ( x  4)2
6 = x+ 4
–1.551 = x
 6= x+ 4
–6.449 = x
Subtract 4 from both sides
x-intercepts at –1.551 and
-6.449
Another example, this time the parabola is concave down. That’s just
another way of saying that it opens downward.
y = –(x – 7)2 + 3
0 = –(x – 7)2 + 3
Subtract 3 from both sides
–3
Divide both sides by -1
= –(x – 7)2
3 = (x – 7)2
Take square root of both sides
3  ( x  7)2
1.732 = x – 7
8.732 = x
–1.732
=x –7
5.268 = x
Add 7 to both sides
x-intercepts at 5.268 and 8.732
Another example, this time the a value is 0.5 (less than 1)
y = 0.5(x + 3)2 + 5
0 = 0.5(x + 3)2 + 5
Subtract 5 from both sides
–5
Divide both sides by 0.5
–10
= 0.5(x + 3)2
= (x + 3)2
10  ( x  3)
Take square root of both sides
2
 10 = x + 3   10= x + 3
An error message will result, because there
is no REAL solution, only an “imaginary”
one.
Subtract 3 from both sides
Thus there are NO x-intercepts.
You cannot take the square root
of a negative number. This is
called a complex solution
involving imaginary numbers.
Alas, you must wait until
Algebra 2!
Find the x-intercepts of the parabola for each quadratic.
1. y = (x – 7)2 – 9
x-intercepts at 10 and 4
2. y = 3(x + 4)2 + 6
NO x-intercepts
3. y = –0.5(x – 2)2 + 10
x-intercepts at 6.472 and –2.472
Is there a way to tell how many x-intercepts a parabola will have
without solving the equation?
Yes, use the discriminant (also called the radicand of the quadratic
formula – see page 293 of textbook)
Finding the y-intercept is a little more straightforward.
Just set x = 0 and solve for y.
y = (x + 4)2 – 6
y = (0 + 4)2 – 6
y = 10
y-intercept at (0, 10)
The quadratic equation does not have to be vertex form
to find the y-intercept.
y = x2 + 8x + 10
y = (0)2 + 8(0) + 10
y = 10
y-intercept at (0, 10)
Complete the Worksheet on Vertex Form of Quadratic Equations