Download Fireworks - From Standard to Vertex Form

Fireworks – From Standard to Vertex Form
• What's the pattern?
(x + 6)2
x
+
6
x
x2
6x
+
6
6x
36
• How about these?
x2 + 4x ______
+ 4
x2 + 10x ______
+ 25
x2 – 14x ______
+ 49



(x _____
+ 2 )2
(x _____
+ 5 )2
(x _____
– 7 )2
x2 + 12x + 36
Fireworks – From Standard to Vertex Form
• Converting 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
Fireworks – From Standard to Vertex Form
• The following equation requires a bit of work to get it
into vertex form.
y = x2 + 8x + 10
y = (x2 + 8x + 16 ) + 10 – 16
82  4
42  16
16 is added to complete
the square. 16 is subtracted to maintain the
balance of the equation.
y = (x + 4)2 – 6
The vertex of this parabola is
located at ( –4, –6 ).
Fireworks – From Standard to Vertex Form
• Lets do another. This time the x2 term is negative.
y = –x2 + 12x – 5
y = (–x2 + 12x
)–5
y = –(x2 – 12x
)–5
Un-distribute a negative
so that when can
complete the square
y = –(x2 – 12x + 36 ) – 5 + 36


12  2 6
(  6) 2  36
The 36 in parentheses
becomes negative so
we must add 36 to keep
the equation balanced.
y = – (x – 6)2 + 31
The vertex of this parabola is
located at ( 6, 31 ).
Fireworks – From Standard to Vertex Form
• 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)
Fireworks – From Standard to Vertex Form
• 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
6 = (x + 4)2
6  ( x  4)
6 = x+ 4
–1.551 = x
2
 6= x+ 4
–6.449 = x
Add 6 to both sides
Take square root of both sides
Subtract 4 from both sides
x-intercepts at –1.551 and
-6.449
Fireworks – From Standard to Vertex Form
• Another example, this time the parabola is concave
down.
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
–1.732
8.732 = x
5.268 = x
=x –7
Add 7 to both sides
x-intercepts at 5.268 and
8.732
Fireworks – From Standard to Vertex Form
• Another example, this time the a value is 0.5.
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)2
 10 = x + 3   10= x + 3
Error = x
Error = x
Take square root of both sides
Subtract 3 from both sides
NO x-intercepts… can't take
square root of a negative number.
Fireworks – From Standard to Vertex Form
• Find the x-intercepts of the parabola for each of the
quadratic equations.
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 doing any calculations?
Fireworks – From Standard to Vertex Form
• 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)