1. Quadratic Equations
Unit 2
English Casbarro

2. Section 1- Graphing Quadratic Equations Quadratic Equations are equations in the form The graphs of quadratic equations are called parabolas.
If a is positive, the parabola points up, ; if a is negative, the parabola points down,
The following graphs show the vertex (–3, 0) of one parabola (which is the only solution), and the solutions (1, 0) and (4, 0) of another parabola. The third parabola has imaginary solutions.

3. Graphing from Standard Form
Standard and Vertex form follow the same
rules. There are 3 things you always find:
The vertex
The y-intercept
The matching point

4. Standard Form: 1. Finding the vertex:
You must use the formula: ,so that you can find the axis of symmetry and the x-value of the vertex.
Next, you substitute the x into the equation to find the y-value of the vertex.
So the vertex is (1,5)

5. Vertex Form:
From the previous slide, we found the vertex of the parabola, using the formula:
and substituting the x value into the equation to find
the y-value of the vertex. You found that the vertex is (1,5).
The “a” from your equation is the same “a” in the “a” in standard form. So,
The equation in vertex form is:

6. Vertex form without completing the square
Now that you can find the vertex of any quadratic
function, it is easy to put the function in vertex form.
Ex. Put in vertex form.
Find So,
Now,
So, the vertex is (1,8) and a = 3.
The equation in vertex form is:

7. Section 7: Quadratic Inequalities
You have graphed in all 3 forms of Quadratic Equations:
Standard Form
Vertex Form
Solutions (factored) Form
Now you will graph the parabolas, then shade for the inequalities.

8. Graphing You must still find the 3 points to graph: The vertex
The y-intercept
The matching point

9. First, we find the vertex:
Example 1:
First, we find the vertex:

10. Next, we find the y-intercept:
Substitute 0 into the inequality:
The y-intercept is (0,3).

11. The matching point The vertex is: The y-intercept is:
Graph these 3 points

12. Shading The inequality was: We always shade “above” the graph for
The equal sign below it means that the line is solid.

13. Graph

14. Using the calculator Go to Y=
Use the direction buttons to move as far to the left as you can
Use the “Enter” button to change the type of display
is represented by the symbol
Type in the quadratic inequality
Graph the inequality: the shading will be automatic.

15. Section 8: Modeling The vertex form of a quadratic equation is:
y + a(x – h)2 + k
You can use this form to find equations if you are given points

16. Given the vertex and a point
Ex. 1: (#10 in your book)
Vertex: (2, –1)
Point: (4, 3)
The equation is y = a(x – h)2 + k
This means that you are given, as the question, 4 out of the 5 variables in the general equation.
(h = 2, k = –1)
(x = 4, y = 3)

17. Fill in the general equation
y = a(x – h)2 + k
3 = a(4 – 2)2 + (–1) (substitute h, k, x, and y)
3 = a(2)2 – 1
3 = 4a – (add 1 to both sides)
4 = 4a (divide both sides by 4)
a = 1

18. Fill in the equation with h, k, a
The general equation is : y = a(x – h)2 + k
(h, k)  (2, –1)
a = 1
So the specific equation is:
y = 1(x – 2)2 – 1

19. You try this:
Vertex: (4, 5)
Point: (8, –3)

20. Solution: y = a(x – h)2 + k Substitute h, k, x, and y
–3 = a(4) Subtract 5 from both sides
–8 = 16a Divide both sides by 16
a = -1/2
So the specific equation is y = –1/2(x – 4)2 + 5