2. Fireworks – From Standard to Vertex Form
• What's the pattern? + x 6 x2 6x 36
(x + 6)2
x2 + 12x + 36
• How about these?
x2 + 4x ______
(x _____ )2
+ 4
+ 2
x x ______
(x _____ )2
+ 25
+ 5
x2 – 14x ______
(x _____ )2
+ 49
– 7

3. Fireworks – From Standard to Vertex Form
• Converting from standard form to vertex form can be easy…
x2 + 6x + 9
(x + 3)2
x2 – 2x =
(x – 1)2
x2 + 8x =
(x + 4)2
x2 + 20x =
(x + 10)2
… but we're not always so lucky

4. 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 ) + 10
+ 16
– 16
16 is added to complete the square. 16 is sub-tracted to maintain the balance of the equation.
y = (x + 4)2 – 6
The vertex of this parabola is located at ( –4, –6 ).

5. Fireworks – From Standard to Vertex Form
• Lets do another. This time the x2 term is negative.
y = –x2 + 12x – 5
Un-distribute a negative so that when can complete the square
y = (–x2 + 12x ) – 5
y = –(x2 – 12x ) – 5
y = –(x2 – 12x ) – 5
+ 36
+ 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 ).

6. 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)

7. 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
Add 6 to both sides
6 = (x + 4)2
Take square root of both sides
= x + 4
= x + 4
Subtract 4 from both sides
–1.551 = x
–6.449 = x
x-intercepts at –1.551 and

8. 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 = –(x – 7)2
Divide both sides by -1
3 = (x – 7)2
Take square root of both sides
1.732 = x – 7
–1.732 = x – 7
Add 7 to both sides
8.732 = x
5.268 = x
x-intercepts at and 8.732

9. 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 = 0.5(x + 3)2
Divide both sides by 0.5
–10 = (x + 3)2
Take square root of both sides
= x + 3
= x + 3
Subtract 3 from both sides
Error = x
Error = x
NO x-intercepts… can't take square root of a negative number.

10. 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 and –2.472
• Is there a way to tell how many x-intercepts a parabola will have without doing any calculations?

11. 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-intercept at (0, 10)
y = 10
• The quadratic equation does not have to be vertex form to find the y-intercept.
y = x2 + 8x + 10
y-intercept at (0, 10)
y = (0)2 + 8(0) + 10
y = 10