1. Vertex Form of Quadratic Equations
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2. General Form
Quadratic functions have the standard form y = ax2 + bx + c
a, b, and c are constants
a ≠ (why?)
Quadratic functions graph as parabolas or U’s

3. 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

4. Axis of Symmetry Parabolas are symmetric about a vertical axis
For y = ax2 + bx + c the axis of symmetry is at
Given y = 3x2 + 8x
What is the axis of symmetry?

5. 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?

6. Vertex of the Parabola Given f(x) = x2 + 2x – 8
What is the x-value of the vertex?
What is the y-value of the vertex?
The vertex is at (-1, -9)

7. 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!

8. 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

9. Vertex Form Changing to vertex form
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)
Now create a
squared binomial
* 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.

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

11. Transforming 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

12. •. The following equation requires a bit of work to get it
• The following equation requires a bit of work to get it into vertex form.
16 is added to both sides in this problem to complete 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.
y = x2 + 8x + 10
= (x2 + 8x )
+ 16
+ 16
y = (x + 4)2 – 6
The vertex is located at ( –4, –6 ) -Be very careful with the signs.

13. • Lets do another. This time the x2 term is negative.
Un-distribute a negative so that you can complete the square
y = – x2 + 12x – 5
y = (– x2 + 12x – 5)
y = – (x2 – 12x )
+ 5
– 5
= – (x2 – 12x )
+ 36
– 5
= – (x2 – 12x )
+ 36
+ 36
+ 31
= – [(x – 6) ]
Bring back the constant and then re-distribute a negative after completing the square.
– 31
= – [(x – 6) ]
+ 31
y = – (x – 6)2 + 31
The vertex of this parabola is located at ( 6, 31 ).

14. • 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)

15. •. In addition to telling us where the vertex is located the
• 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
Subtract 4 from both sides
= x + 4
= x + 4
–1.551 = x
–6.449 = x
x-intercepts at –1.551 and

16. y = –(x – 7)2 + 3 0 = –(x – 7)2 + 3 –3 = –(x – 7)2 3 = (x – 7)2
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 = –(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

17. 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 = 0.5(x + 3)2
Divide both sides by 0.5
–10 = (x + 3)2
Take square root of both sides
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!
= x + 3
= x + 3
An error message will result, because there is no REAL solution, only an “imaginary” one.

18. 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 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)

19. Finding the y-intercept is a little more straightforward.
Just set x = 0 and solve for y.
y = (x + 4)2 – 6
y-intercept at (0, 10)
y = (0 + 4)2 – 6
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

20. HOMEWORK
Complete the Worksheet on Vertex Form of Quadratic Equations